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Given the equation $-3 x^2+3 x-9 y^2+9 y+6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-9 y^2+9 y-3 x^2+3 x+6=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }6 \text{from }\text{both }\text{sides}: \\
-9 y^2+9 y-3 x^2+3 x=-6 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-3 x^2+3 x+\underline{\text{ }}\right)+\left(-9 y^2+9 y+\underline{\text{ }}\right)=\underline{\text{ }}-6 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-3 x^2+3 x+\underline{\text{ }}\right)=-3 \left(x^2-x+\underline{\text{ }}\right): \\
\fbox{$-3 \left(x^2-x+\underline{\text{ }}\right)$}+\left(-9 y^2+9 y+\underline{\text{ }}\right)=\underline{\text{ }}-6 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-9 y^2+9 y+\underline{\text{ }}\right)=-9 \left(y^2-y+\underline{\text{ }}\right): \\
-3 \left(x^2-x+\underline{\text{ }}\right)+\fbox{$-9 \left(y^2-y+\underline{\text{ }}\right)$}=\underline{\text{ }}-6 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{-3}{4}=-\frac{3}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-6-\frac{3}{4}=-\frac{27}{4}: \\
-3 \left(x^2-x+\frac{1}{4}\right)-9 \left(y^2-y+\underline{\text{ }}\right)=\fbox{$-\frac{27}{4}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{-9}{4}=-\frac{9}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{27}{4}-\frac{9}{4}=-9: \\
-3 \left(x^2-x+\frac{1}{4}\right)-9 \left(y^2-y+\frac{1}{4}\right)=\fbox{$-9$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-x+\frac{1}{4}=\left(x-\frac{1}{2}\right)^2: \\
-3 \fbox{$\left(x-\frac{1}{2}\right)^2$}-9 \left(y^2-y+\frac{1}{4}\right)=-9 \\
\end{array}
Step 11:
\begin{array}{l}
y^2-y+\frac{1}{4}=\left(y-\frac{1}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -3 \left(x-\frac{1}{2}\right)^2-9 \fbox{$\left(y-\frac{1}{2}\right)^2$}=-9 \\
\end{array}
| khanacademy | amps |
Given the equation $2 x^2+9 x+y^2+2 y+8=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
y^2+2 y+2 x^2+9 x+8=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }8 \text{from }\text{both }\text{sides}: \\
y^2+2 y+2 x^2+9 x=-8 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(2 x^2+9 x+\underline{\text{ }}\right)+\left(y^2+2 y+\underline{\text{ }}\right)=\underline{\text{ }}-8 \\
\end{array}
Step 4:
\begin{array}{l}
\left(2 x^2+9 x+\underline{\text{ }}\right)=2 \left(x^2+\frac{9 x}{2}+\underline{\text{ }}\right): \\
\fbox{$2 \left(x^2+\frac{9 x}{2}+\underline{\text{ }}\right)$}+\left(y^2+2 y+\underline{\text{ }}\right)=\underline{\text{ }}-8 \\
\end{array}
Step 5:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{9}{2}}{2}\right)^2=\frac{81}{16} \text{on }\text{the }\text{left }\text{and }2\times \frac{81}{16}=\frac{81}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 6:
\begin{array}{l}
\frac{81}{8}-8=\frac{17}{8}: \\
2 \left(x^2+\frac{9 x}{2}+\frac{81}{16}\right)+\left(y^2+2 y+\underline{\text{ }}\right)=\fbox{$\frac{17}{8}$} \\
\end{array}
Step 7:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it, }\text{then }\text{add }\text{it }\text{to }\text{both }\text{sides.} \\
\text{Add }\left(\frac{2}{2}\right)^2=1 \text{to }\text{both }\text{sides}: \\
\end{array}
Step 8:
\begin{array}{l}
\frac{17}{8}+1=\frac{25}{8}: \\
2 \left(x^2+\frac{9 x}{2}+\frac{81}{16}\right)+\left(y^2+2 y+1\right)=\fbox{$\frac{25}{8}$} \\
\end{array}
Step 9:
\begin{array}{l}
x^2+\frac{9 x}{2}+\frac{81}{16}=\left(x+\frac{9}{4}\right)^2: \\
2 \fbox{$\left(x+\frac{9}{4}\right)^2$}+\left(y^2+2 y+1\right)=\frac{25}{8} \\
\end{array}
Step 10:
\begin{array}{l}
y^2+2 y+1=(y+1)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 2 \left(x+\frac{9}{4}\right)^2+\fbox{$(y+1)^2$}=\frac{25}{8} \\
\end{array}
| khanacademy | amps |
Given the equation $4 x^2+9 x+2 y^2-8 y-5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
2 y^2-8 y+4 x^2+9 x-5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }5 \text{to }\text{both }\text{sides}: \\
2 y^2-8 y+4 x^2+9 x=5 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(4 x^2+9 x+\underline{\text{ }}\right)+\left(2 y^2-8 y+\underline{\text{ }}\right)=\underline{\text{ }}+5 \\
\end{array}
Step 4:
\begin{array}{l}
\left(4 x^2+9 x+\underline{\text{ }}\right)=4 \left(x^2+\frac{9 x}{4}+\underline{\text{ }}\right): \\
\fbox{$4 \left(x^2+\frac{9 x}{4}+\underline{\text{ }}\right)$}+\left(2 y^2-8 y+\underline{\text{ }}\right)=\underline{\text{ }}+5 \\
\end{array}
Step 5:
\begin{array}{l}
\left(2 y^2-8 y+\underline{\text{ }}\right)=2 \left(y^2-4 y+\underline{\text{ }}\right): \\
4 \left(x^2+\frac{9 x}{4}+\underline{\text{ }}\right)+\fbox{$2 \left(y^2-4 y+\underline{\text{ }}\right)$}=\underline{\text{ }}+5 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{9}{4}}{2}\right)^2=\frac{81}{64} \text{on }\text{the }\text{left }\text{and }4\times \frac{81}{64}=\frac{81}{16} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
5+\frac{81}{16}=\frac{161}{16}: \\
4 \left(x^2+\frac{9 x}{4}+\frac{81}{64}\right)+2 \left(y^2-4 y+\underline{\text{ }}\right)=\fbox{$\frac{161}{16}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-4}{2}\right)^2=4 \text{on }\text{the }\text{left }\text{and }2\times 4=8 \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{161}{16}+8=\frac{289}{16}: \\
4 \left(x^2+\frac{9 x}{4}+\frac{81}{64}\right)+2 \left(y^2-4 y+4\right)=\fbox{$\frac{289}{16}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{9 x}{4}+\frac{81}{64}=\left(x+\frac{9}{8}\right)^2: \\
4 \fbox{$\left(x+\frac{9}{8}\right)^2$}+2 \left(y^2-4 y+4\right)=\frac{289}{16} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-4 y+4=(y-2)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 4 \left(x+\frac{9}{8}\right)^2+2 \fbox{$(y-2)^2$}=\frac{289}{16} \\
\end{array}
| khanacademy | amps |
Given the equation $6 x^2+8 x-y^2+10 y-6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-y^2+10 y+6 x^2+8 x-6=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }6 \text{to }\text{both }\text{sides}: \\
-y^2+10 y+6 x^2+8 x=6 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(6 x^2+8 x+\underline{\text{ }}\right)+\left(-y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\end{array}
Step 4:
\begin{array}{l}
\left(6 x^2+8 x+\underline{\text{ }}\right)=6 \left(x^2+\frac{4 x}{3}+\underline{\text{ }}\right): \\
\fbox{$6 \left(x^2+\frac{4 x}{3}+\underline{\text{ }}\right)$}+\left(-y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-y^2+10 y+\underline{\text{ }}\right)=-\left(y^2-10 y+\underline{\text{ }}\right): \\
6 \left(x^2+\frac{4 x}{3}+\underline{\text{ }}\right)+\fbox{$-\left(y^2-10 y+\underline{\text{ }}\right)$}=\underline{\text{ }}+6 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{4}{3}}{2}\right)^2=\frac{4}{9} \text{on }\text{the }\text{left }\text{and }6\times \frac{4}{9}=\frac{8}{3} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
6+\frac{8}{3}=\frac{26}{3}: \\
6 \left(x^2+\frac{4 x}{3}+\frac{4}{9}\right)-\left(y^2-10 y+\underline{\text{ }}\right)=\fbox{$\frac{26}{3}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-1 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-10}{2}\right)^2=25 \text{on }\text{the }\text{left }\text{and }-25=-25 \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{26}{3}-25=-\frac{49}{3}: \\
6 \left(x^2+\frac{4 x}{3}+\frac{4}{9}\right)-\left(y^2-10 y+25\right)=\fbox{$-\frac{49}{3}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{4 x}{3}+\frac{4}{9}=\left(x+\frac{2}{3}\right)^2: \\
6 \fbox{$\left(x+\frac{2}{3}\right)^2$}-\left(y^2-10 y+25\right)=-\frac{49}{3} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-10 y+25=(y-5)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 6 \left(x+\frac{2}{3}\right)^2-\fbox{$(y-5)^2$}=-\frac{49}{3} \\
\end{array}
| khanacademy | amps |
Given the equation $5 x^2-3 x+2 y^2-8 y+4=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
2 y^2-8 y+5 x^2-3 x+4=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }4 \text{from }\text{both }\text{sides}: \\
2 y^2-8 y+5 x^2-3 x=-4 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(5 x^2-3 x+\underline{\text{ }}\right)+\left(2 y^2-8 y+\underline{\text{ }}\right)=\underline{\text{ }}-4 \\
\end{array}
Step 4:
\begin{array}{l}
\left(5 x^2-3 x+\underline{\text{ }}\right)=5 \left(x^2-\frac{3 x}{5}+\underline{\text{ }}\right): \\
\fbox{$5 \left(x^2-\frac{3 x}{5}+\underline{\text{ }}\right)$}+\left(2 y^2-8 y+\underline{\text{ }}\right)=\underline{\text{ }}-4 \\
\end{array}
Step 5:
\begin{array}{l}
\left(2 y^2-8 y+\underline{\text{ }}\right)=2 \left(y^2-4 y+\underline{\text{ }}\right): \\
5 \left(x^2-\frac{3 x}{5}+\underline{\text{ }}\right)+\fbox{$2 \left(y^2-4 y+\underline{\text{ }}\right)$}=\underline{\text{ }}-4 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-3}{5}}{2}\right)^2=\frac{9}{100} \text{on }\text{the }\text{left }\text{and }5\times \frac{9}{100}=\frac{9}{20} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{9}{20}-4=-\frac{71}{20}: \\
5 \left(x^2-\frac{3 x}{5}+\frac{9}{100}\right)+2 \left(y^2-4 y+\underline{\text{ }}\right)=\fbox{$-\frac{71}{20}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-4}{2}\right)^2=4 \text{on }\text{the }\text{left }\text{and }2\times 4=8 \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
8-\frac{71}{20}=\frac{89}{20}: \\
5 \left(x^2-\frac{3 x}{5}+\frac{9}{100}\right)+2 \left(y^2-4 y+4\right)=\fbox{$\frac{89}{20}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{3 x}{5}+\frac{9}{100}=\left(x-\frac{3}{10}\right)^2: \\
5 \fbox{$\left(x-\frac{3}{10}\right)^2$}+2 \left(y^2-4 y+4\right)=\frac{89}{20} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-4 y+4=(y-2)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 5 \left(x-\frac{3}{10}\right)^2+2 \fbox{$(y-2)^2$}=\frac{89}{20} \\
\end{array}
| khanacademy | amps |
Given the equation $10 x^2-7 x+7 y-8=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
10 x^2-7 x+(7 y-8)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }7 y-8 \text{from }\text{both }\text{sides}: \\
10 x^2-7 x=8-7 y \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{together }\text{on }\text{the }\text{left }\text{hand }\text{side, }\text{with }\text{a }\text{placeholder }\text{constant}: \\
\left(10 x^2-7 x+\underline{\text{ }}\right)=(8-7 y)+\underline{\text{ }} \\
\end{array}
Step 4:
\begin{array}{l}
\left(10 x^2-7 x+\underline{\text{ }}\right)=10 \left(x^2-\frac{7 x}{10}+\underline{\text{ }}\right): \\
\fbox{$10 \left(x^2-\frac{7 x}{10}+\underline{\text{ }}\right)$}=(8-7 y)+\underline{\text{ }} \\
\end{array}
Step 5:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }\text{10 }\text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-7}{10}}{2}\right)^2=\frac{49}{400} \text{on }\text{the }\text{left }\text{and }10\times \frac{49}{400}=\frac{49}{40} \text{on }\text{the }\text{right}: \\
\end{array}
Step 6:
\begin{array}{l}
(8-7 y)+\frac{49}{40}=\frac{369}{40}-7 y: \\
10 \left(x^2-\frac{7 x}{10}+\frac{49}{400}\right)=\fbox{$\frac{369}{40}-7 y$} \\
\end{array}
Step 7:
\begin{array}{l}
x^2-\frac{7 x}{10}+\frac{49}{400}=\left(x-\frac{7}{20}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & \text{10 }\fbox{$\left(x-\frac{7}{20}\right)^2$}=\frac{369}{40}-7 y \\
\end{array}
| khanacademy | amps |
Given the equation $10 x^2+9 x-6 y^2-3 y+7=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-6 y^2-3 y+10 x^2+9 x+7=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }7 \text{from }\text{both }\text{sides}: \\
-6 y^2-3 y+10 x^2+9 x=-7 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(10 x^2+9 x+\underline{\text{ }}\right)+\left(-6 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}-7 \\
\end{array}
Step 4:
\begin{array}{l}
\left(10 x^2+9 x+\underline{\text{ }}\right)=10 \left(x^2+\frac{9 x}{10}+\underline{\text{ }}\right): \\
\fbox{$10 \left(x^2+\frac{9 x}{10}+\underline{\text{ }}\right)$}+\left(-6 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}-7 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-6 y^2-3 y+\underline{\text{ }}\right)=-6 \left(y^2+\frac{y}{2}+\underline{\text{ }}\right): \\
10 \left(x^2+\frac{9 x}{10}+\underline{\text{ }}\right)+\fbox{$-6 \left(y^2+\frac{y}{2}+\underline{\text{ }}\right)$}=\underline{\text{ }}-7 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }\text{10 }\text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{9}{10}}{2}\right)^2=\frac{81}{400} \text{on }\text{the }\text{left }\text{and }10\times \frac{81}{400}=\frac{81}{40} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{81}{40}-7=-\frac{199}{40}: \\
10 \left(x^2+\frac{9 x}{10}+\frac{81}{400}\right)-6 \left(y^2+\frac{y}{2}+\underline{\text{ }}\right)=\fbox{$-\frac{199}{40}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{2}}{2}\right)^2=\frac{1}{16} \text{on }\text{the }\text{left }\text{and }\frac{-6}{16}=-\frac{3}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{199}{40}-\frac{3}{8}=-\frac{107}{20}: \\
10 \left(x^2+\frac{9 x}{10}+\frac{81}{400}\right)-6 \left(y^2+\frac{y}{2}+\frac{1}{16}\right)=\fbox{$-\frac{107}{20}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{9 x}{10}+\frac{81}{400}=\left(x+\frac{9}{20}\right)^2: \\
\text{10 }\fbox{$\left(x+\frac{9}{20}\right)^2$}-6 \left(y^2+\frac{y}{2}+\frac{1}{16}\right)=-\frac{107}{20} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{y}{2}+\frac{1}{16}=\left(y+\frac{1}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 10 \left(x+\frac{9}{20}\right)^2-6 \fbox{$\left(y+\frac{1}{4}\right)^2$}=-\frac{107}{20} \\
\end{array}
| khanacademy | amps |
Given the equation $-x^2-8 x-4 y^2-y+9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-4 y^2-y-x^2-8 x+9=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }9 \text{from }\text{both }\text{sides}: \\
-4 y^2-y-x^2-8 x=-9 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-x^2-8 x+\underline{\text{ }}\right)+\left(-4 y^2-y+\underline{\text{ }}\right)=\underline{\text{ }}-9 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-x^2-8 x+\underline{\text{ }}\right)=-\left(x^2+8 x+\underline{\text{ }}\right): \\
\fbox{$-\left(x^2+8 x+\underline{\text{ }}\right)$}+\left(-4 y^2-y+\underline{\text{ }}\right)=\underline{\text{ }}-9 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-4 y^2-y+\underline{\text{ }}\right)=-4 \left(y^2+\frac{y}{4}+\underline{\text{ }}\right): \\
-\left(x^2+8 x+\underline{\text{ }}\right)+\fbox{$-4 \left(y^2+\frac{y}{4}+\underline{\text{ }}\right)$}=\underline{\text{ }}-9 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-1 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{8}{2}\right)^2=16 \text{on }\text{the }\text{left }\text{and }-16=-16 \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-9-16=-25: \\
-\left(x^2+8 x+16\right)-4 \left(y^2+\frac{y}{4}+\underline{\text{ }}\right)=\fbox{$-25$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{4}}{2}\right)^2=\frac{1}{64} \text{on }\text{the }\text{left }\text{and }\frac{-4}{64}=-\frac{1}{16} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-25-\frac{1}{16}=-\frac{401}{16}: \\
-\left(x^2+8 x+16\right)-4 \left(y^2+\frac{y}{4}+\frac{1}{64}\right)=\fbox{$-\frac{401}{16}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+8 x+16=(x+4)^2: \\
-\fbox{$(x+4)^2$}-4 \left(y^2+\frac{y}{4}+\frac{1}{64}\right)=-\frac{401}{16} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{y}{4}+\frac{1}{64}=\left(y+\frac{1}{8}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -(x+4)^2-4 \fbox{$\left(y+\frac{1}{8}\right)^2$}=-\frac{401}{16} \\
\end{array}
| khanacademy | amps |
Given the equation $7 x^2+5 x-4 y^2-y-3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-4 y^2-y+7 x^2+5 x-3=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }3 \text{to }\text{both }\text{sides}: \\
-4 y^2-y+7 x^2+5 x=3 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(7 x^2+5 x+\underline{\text{ }}\right)+\left(-4 y^2-y+\underline{\text{ }}\right)=\underline{\text{ }}+3 \\
\end{array}
Step 4:
\begin{array}{l}
\left(7 x^2+5 x+\underline{\text{ }}\right)=7 \left(x^2+\frac{5 x}{7}+\underline{\text{ }}\right): \\
\fbox{$7 \left(x^2+\frac{5 x}{7}+\underline{\text{ }}\right)$}+\left(-4 y^2-y+\underline{\text{ }}\right)=\underline{\text{ }}+3 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-4 y^2-y+\underline{\text{ }}\right)=-4 \left(y^2+\frac{y}{4}+\underline{\text{ }}\right): \\
7 \left(x^2+\frac{5 x}{7}+\underline{\text{ }}\right)+\fbox{$-4 \left(y^2+\frac{y}{4}+\underline{\text{ }}\right)$}=\underline{\text{ }}+3 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{5}{7}}{2}\right)^2=\frac{25}{196} \text{on }\text{the }\text{left }\text{and }7\times \frac{25}{196}=\frac{25}{28} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
3+\frac{25}{28}=\frac{109}{28}: \\
7 \left(x^2+\frac{5 x}{7}+\frac{25}{196}\right)-4 \left(y^2+\frac{y}{4}+\underline{\text{ }}\right)=\fbox{$\frac{109}{28}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{4}}{2}\right)^2=\frac{1}{64} \text{on }\text{the }\text{left }\text{and }\frac{-4}{64}=-\frac{1}{16} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{109}{28}-\frac{1}{16}=\frac{429}{112}: \\
7 \left(x^2+\frac{5 x}{7}+\frac{25}{196}\right)-4 \left(y^2+\frac{y}{4}+\frac{1}{64}\right)=\fbox{$\frac{429}{112}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{5 x}{7}+\frac{25}{196}=\left(x+\frac{5}{14}\right)^2: \\
7 \fbox{$\left(x+\frac{5}{14}\right)^2$}-4 \left(y^2+\frac{y}{4}+\frac{1}{64}\right)=\frac{429}{112} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{y}{4}+\frac{1}{64}=\left(y+\frac{1}{8}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 7 \left(x+\frac{5}{14}\right)^2-4 \fbox{$\left(y+\frac{1}{8}\right)^2$}=\frac{429}{112} \\
\end{array}
| khanacademy | amps |
Given the equation $-x^2+8 x-4 y^2+7 y-5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-4 y^2+7 y-x^2+8 x-5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }5 \text{to }\text{both }\text{sides}: \\
-4 y^2+7 y-x^2+8 x=5 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-x^2+8 x+\underline{\text{ }}\right)+\left(-4 y^2+7 y+\underline{\text{ }}\right)=\underline{\text{ }}+5 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-x^2+8 x+\underline{\text{ }}\right)=-\left(x^2-8 x+\underline{\text{ }}\right): \\
\fbox{$-\left(x^2-8 x+\underline{\text{ }}\right)$}+\left(-4 y^2+7 y+\underline{\text{ }}\right)=\underline{\text{ }}+5 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-4 y^2+7 y+\underline{\text{ }}\right)=-4 \left(y^2-\frac{7 y}{4}+\underline{\text{ }}\right): \\
-\left(x^2-8 x+\underline{\text{ }}\right)+\fbox{$-4 \left(y^2-\frac{7 y}{4}+\underline{\text{ }}\right)$}=\underline{\text{ }}+5 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-1 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-8}{2}\right)^2=16 \text{on }\text{the }\text{left }\text{and }-16=-16 \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
5-16=-11: \\
-\left(x^2-8 x+16\right)-4 \left(y^2-\frac{7 y}{4}+\underline{\text{ }}\right)=\fbox{$-11$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-7}{4}}{2}\right)^2=\frac{49}{64} \text{on }\text{the }\text{left }\text{and }-4\times \frac{49}{64}=-\frac{49}{16} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-11-\frac{49}{16}=-\frac{225}{16}: \\
-\left(x^2-8 x+16\right)-4 \left(y^2-\frac{7 y}{4}+\frac{49}{64}\right)=\fbox{$-\frac{225}{16}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-8 x+16=(x-4)^2: \\
-\fbox{$(x-4)^2$}-4 \left(y^2-\frac{7 y}{4}+\frac{49}{64}\right)=-\frac{225}{16} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{7 y}{4}+\frac{49}{64}=\left(y-\frac{7}{8}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -(x-4)^2-4 \fbox{$\left(y-\frac{7}{8}\right)^2$}=-\frac{225}{16} \\
\end{array}
| khanacademy | amps |
Given the equation $10 x^2+7 y^2-5 y+9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
7 y^2-5 y+\left(10 x^2+9\right)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }9 \text{from }\text{both }\text{sides}: \\
7 y^2-5 y+10 x^2=-9 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{together }\text{on }\text{the }\text{left }\text{hand }\text{side, }\text{with }\text{a }\text{placeholder }\text{constant}: \\
\left(7 y^2-5 y+\underline{\text{ }}\right)+10 x^2=\underline{\text{ }}-9 \\
\end{array}
Step 4:
\begin{array}{l}
\left(7 y^2-5 y+\underline{\text{ }}\right)=7 \left(y^2-\frac{5 y}{7}+\underline{\text{ }}\right): \\
\fbox{$7 \left(y^2-\frac{5 y}{7}+\underline{\text{ }}\right)$}+10 x^2=\underline{\text{ }}-9 \\
\end{array}
Step 5:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-5}{7}}{2}\right)^2=\frac{25}{196} \text{on }\text{the }\text{left }\text{and }7\times \frac{25}{196}=\frac{25}{28} \text{on }\text{the }\text{right}: \\
\end{array}
Step 6:
\begin{array}{l}
\frac{25}{28}-9=-\frac{227}{28}: \\
7 \left(y^2-\frac{5 y}{7}+\frac{25}{196}\right)+10 x^2=\fbox{$-\frac{227}{28}$} \\
\end{array}
Step 7:
\begin{array}{l}
y^2-\frac{5 y}{7}+\frac{25}{196}=\left(y-\frac{5}{14}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 7 \fbox{$\left(y-\frac{5}{14}\right)^2$}+10 x^2=-\frac{227}{28} \\
\end{array}
| khanacademy | amps |
Given the equation $-9 x^2-8 y^2-y-3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-8 y^2-y+\left(-9 x^2-3\right)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }8 y^2+y+9 x^2+3 \text{to }\text{both }\text{sides}: \\
8 y^2+y+\left(9 x^2+3\right)=0 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Subtract }3 \text{from }\text{both }\text{sides}: \\
8 y^2+y+9 x^2=-3 \\
\end{array}
Step 4:
\begin{array}{l}
\text{Group }\text{terms }\text{together }\text{on }\text{the }\text{left }\text{hand }\text{side, }\text{with }\text{a }\text{placeholder }\text{constant}: \\
\left(8 y^2+y+\underline{\text{ }}\right)+9 x^2=\underline{\text{ }}-3 \\
\end{array}
Step 5:
\begin{array}{l}
\left(8 y^2+y+\underline{\text{ }}\right)=8 \left(y^2+\frac{y}{8}+\underline{\text{ }}\right): \\
\fbox{$8 \left(y^2+\frac{y}{8}+\underline{\text{ }}\right)$}+9 x^2=\underline{\text{ }}-3 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{8}}{2}\right)^2=\frac{1}{256} \text{on }\text{the }\text{left }\text{and }\frac{8}{256}=\frac{1}{32} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{1}{32}-3=-\frac{95}{32}: \\
8 \left(y^2+\frac{y}{8}+\frac{1}{256}\right)+9 x^2=\fbox{$-\frac{95}{32}$} \\
\end{array}
Step 8:
\begin{array}{l}
y^2+\frac{y}{8}+\frac{1}{256}=\left(y+\frac{1}{16}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 8 \fbox{$\left(y+\frac{1}{16}\right)^2$}+9 x^2=-\frac{95}{32} \\
\end{array}
| khanacademy | amps |
Given the equation $-8 x^2+5 x+y^2-7 y+5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
y^2-7 y-8 x^2+5 x+5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }5 \text{from }\text{both }\text{sides}: \\
y^2-7 y-8 x^2+5 x=-5 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-8 x^2+5 x+\underline{\text{ }}\right)+\left(y^2-7 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-8 x^2+5 x+\underline{\text{ }}\right)=-8 \left(x^2-\frac{5 x}{8}+\underline{\text{ }}\right): \\
\fbox{$-8 \left(x^2-\frac{5 x}{8}+\underline{\text{ }}\right)$}+\left(y^2-7 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\end{array}
Step 5:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-5}{8}}{2}\right)^2=\frac{25}{256} \text{on }\text{the }\text{left }\text{and }-8\times \frac{25}{256}=-\frac{25}{32} \text{on }\text{the }\text{right}: \\
\end{array}
Step 6:
\begin{array}{l}
-5-\frac{25}{32}=-\frac{185}{32}: \\
-8 \left(x^2-\frac{5 x}{8}+\frac{25}{256}\right)+\left(y^2-7 y+\underline{\text{ }}\right)=\fbox{$-\frac{185}{32}$} \\
\end{array}
Step 7:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it, }\text{then }\text{add }\text{it }\text{to }\text{both }\text{sides.} \\
\text{Add }\left(\frac{-7}{2}\right)^2=\frac{49}{4} \text{to }\text{both }\text{sides}: \\
\end{array}
Step 8:
\begin{array}{l}
\frac{49}{4}-\frac{185}{32}=\frac{207}{32}: \\
-8 \left(x^2-\frac{5 x}{8}+\frac{25}{256}\right)+\left(y^2-7 y+\frac{49}{4}\right)=\fbox{$\frac{207}{32}$} \\
\end{array}
Step 9:
\begin{array}{l}
x^2-\frac{5 x}{8}+\frac{25}{256}=\left(x-\frac{5}{16}\right)^2: \\
-8 \fbox{$\left(x-\frac{5}{16}\right)^2$}+\left(y^2-7 y+\frac{49}{4}\right)=\frac{207}{32} \\
\end{array}
Step 10:
\begin{array}{l}
y^2-7 y+\frac{49}{4}=\left(y-\frac{7}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -8 \left(x-\frac{5}{16}\right)^2+\fbox{$\left(y-\frac{7}{2}\right)^2$}=\frac{207}{32} \\
\end{array}
| khanacademy | amps |
Given the equation $6 x^2+9 x+3 y^2-8 y+2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
3 y^2-8 y+6 x^2+9 x+2=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }2 \text{from }\text{both }\text{sides}: \\
3 y^2-8 y+6 x^2+9 x=-2 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(6 x^2+9 x+\underline{\text{ }}\right)+\left(3 y^2-8 y+\underline{\text{ }}\right)=\underline{\text{ }}-2 \\
\end{array}
Step 4:
\begin{array}{l}
\left(6 x^2+9 x+\underline{\text{ }}\right)=6 \left(x^2+\frac{3 x}{2}+\underline{\text{ }}\right): \\
\fbox{$6 \left(x^2+\frac{3 x}{2}+\underline{\text{ }}\right)$}+\left(3 y^2-8 y+\underline{\text{ }}\right)=\underline{\text{ }}-2 \\
\end{array}
Step 5:
\begin{array}{l}
\left(3 y^2-8 y+\underline{\text{ }}\right)=3 \left(y^2-\frac{8 y}{3}+\underline{\text{ }}\right): \\
6 \left(x^2+\frac{3 x}{2}+\underline{\text{ }}\right)+\fbox{$3 \left(y^2-\frac{8 y}{3}+\underline{\text{ }}\right)$}=\underline{\text{ }}-2 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{3}{2}}{2}\right)^2=\frac{9}{16} \text{on }\text{the }\text{left }\text{and }6\times \frac{9}{16}=\frac{27}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{27}{8}-2=\frac{11}{8}: \\
6 \left(x^2+\frac{3 x}{2}+\frac{9}{16}\right)+3 \left(y^2-\frac{8 y}{3}+\underline{\text{ }}\right)=\fbox{$\frac{11}{8}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-8}{3}}{2}\right)^2=\frac{16}{9} \text{on }\text{the }\text{left }\text{and }3\times \frac{16}{9}=\frac{16}{3} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{11}{8}+\frac{16}{3}=\frac{161}{24}: \\
6 \left(x^2+\frac{3 x}{2}+\frac{9}{16}\right)+3 \left(y^2-\frac{8 y}{3}+\frac{16}{9}\right)=\fbox{$\frac{161}{24}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{3 x}{2}+\frac{9}{16}=\left(x+\frac{3}{4}\right)^2: \\
6 \fbox{$\left(x+\frac{3}{4}\right)^2$}+3 \left(y^2-\frac{8 y}{3}+\frac{16}{9}\right)=\frac{161}{24} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{8 y}{3}+\frac{16}{9}=\left(y-\frac{4}{3}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 6 \left(x+\frac{3}{4}\right)^2+3 \fbox{$\left(y-\frac{4}{3}\right)^2$}=\frac{161}{24} \\
\end{array}
| khanacademy | amps |
Given the equation $-6 x^2-3 x+6 y^2+y-9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
6 y^2+y-6 x^2-3 x-9=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }9 \text{to }\text{both }\text{sides}: \\
6 y^2+y-6 x^2-3 x=9 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-6 x^2-3 x+\underline{\text{ }}\right)+\left(6 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}+9 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-6 x^2-3 x+\underline{\text{ }}\right)=-6 \left(x^2+\frac{x}{2}+\underline{\text{ }}\right): \\
\fbox{$-6 \left(x^2+\frac{x}{2}+\underline{\text{ }}\right)$}+\left(6 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}+9 \\
\end{array}
Step 5:
\begin{array}{l}
\left(6 y^2+y+\underline{\text{ }}\right)=6 \left(y^2+\frac{y}{6}+\underline{\text{ }}\right): \\
-6 \left(x^2+\frac{x}{2}+\underline{\text{ }}\right)+\fbox{$6 \left(y^2+\frac{y}{6}+\underline{\text{ }}\right)$}=\underline{\text{ }}+9 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{2}}{2}\right)^2=\frac{1}{16} \text{on }\text{the }\text{left }\text{and }\frac{-6}{16}=-\frac{3}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
9-\frac{3}{8}=\frac{69}{8}: \\
-6 \left(x^2+\frac{x}{2}+\frac{1}{16}\right)+6 \left(y^2+\frac{y}{6}+\underline{\text{ }}\right)=\fbox{$\frac{69}{8}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{6}}{2}\right)^2=\frac{1}{144} \text{on }\text{the }\text{left }\text{and }\frac{6}{144}=\frac{1}{24} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{69}{8}+\frac{1}{24}=\frac{26}{3}: \\
-6 \left(x^2+\frac{x}{2}+\frac{1}{16}\right)+6 \left(y^2+\frac{y}{6}+\frac{1}{144}\right)=\fbox{$\frac{26}{3}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{x}{2}+\frac{1}{16}=\left(x+\frac{1}{4}\right)^2: \\
-6 \fbox{$\left(x+\frac{1}{4}\right)^2$}+6 \left(y^2+\frac{y}{6}+\frac{1}{144}\right)=\frac{26}{3} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{y}{6}+\frac{1}{144}=\left(y+\frac{1}{12}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -6 \left(x+\frac{1}{4}\right)^2+6 \fbox{$\left(y+\frac{1}{12}\right)^2$}=\frac{26}{3} \\
\end{array}
| khanacademy | amps |
Given the equation $-9 x^2-8 x-2 y^2+9 y-4=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-2 y^2+9 y-9 x^2-8 x-4=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }4 \text{to }\text{both }\text{sides}: \\
-2 y^2+9 y-9 x^2-8 x=4 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-9 x^2-8 x+\underline{\text{ }}\right)+\left(-2 y^2+9 y+\underline{\text{ }}\right)=\underline{\text{ }}+4 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-9 x^2-8 x+\underline{\text{ }}\right)=-9 \left(x^2+\frac{8 x}{9}+\underline{\text{ }}\right): \\
\fbox{$-9 \left(x^2+\frac{8 x}{9}+\underline{\text{ }}\right)$}+\left(-2 y^2+9 y+\underline{\text{ }}\right)=\underline{\text{ }}+4 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-2 y^2+9 y+\underline{\text{ }}\right)=-2 \left(y^2-\frac{9 y}{2}+\underline{\text{ }}\right): \\
-9 \left(x^2+\frac{8 x}{9}+\underline{\text{ }}\right)+\fbox{$-2 \left(y^2-\frac{9 y}{2}+\underline{\text{ }}\right)$}=\underline{\text{ }}+4 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{8}{9}}{2}\right)^2=\frac{16}{81} \text{on }\text{the }\text{left }\text{and }-9\times \frac{16}{81}=-\frac{16}{9} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
4-\frac{16}{9}=\frac{20}{9}: \\
-9 \left(x^2+\frac{8 x}{9}+\frac{16}{81}\right)-2 \left(y^2-\frac{9 y}{2}+\underline{\text{ }}\right)=\fbox{$\frac{20}{9}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-9}{2}}{2}\right)^2=\frac{81}{16} \text{on }\text{the }\text{left }\text{and }-2\times \frac{81}{16}=-\frac{81}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{20}{9}-\frac{81}{8}=-\frac{569}{72}: \\
-9 \left(x^2+\frac{8 x}{9}+\frac{16}{81}\right)-2 \left(y^2-\frac{9 y}{2}+\frac{81}{16}\right)=\fbox{$-\frac{569}{72}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{8 x}{9}+\frac{16}{81}=\left(x+\frac{4}{9}\right)^2: \\
-9 \fbox{$\left(x+\frac{4}{9}\right)^2$}-2 \left(y^2-\frac{9 y}{2}+\frac{81}{16}\right)=-\frac{569}{72} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{9 y}{2}+\frac{81}{16}=\left(y-\frac{9}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -9 \left(x+\frac{4}{9}\right)^2-2 \fbox{$\left(y-\frac{9}{4}\right)^2$}=-\frac{569}{72} \\
\end{array}
| khanacademy | amps |
Given the equation $7 x^2-6 x-6 y^2+y-2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-6 y^2+y+7 x^2-6 x-2=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }2 \text{to }\text{both }\text{sides}: \\
-6 y^2+y+7 x^2-6 x=2 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(7 x^2-6 x+\underline{\text{ }}\right)+\left(-6 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}+2 \\
\end{array}
Step 4:
\begin{array}{l}
\left(7 x^2-6 x+\underline{\text{ }}\right)=7 \left(x^2-\frac{6 x}{7}+\underline{\text{ }}\right): \\
\fbox{$7 \left(x^2-\frac{6 x}{7}+\underline{\text{ }}\right)$}+\left(-6 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}+2 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-6 y^2+y+\underline{\text{ }}\right)=-6 \left(y^2-\frac{y}{6}+\underline{\text{ }}\right): \\
7 \left(x^2-\frac{6 x}{7}+\underline{\text{ }}\right)+\fbox{$-6 \left(y^2-\frac{y}{6}+\underline{\text{ }}\right)$}=\underline{\text{ }}+2 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-6}{7}}{2}\right)^2=\frac{9}{49} \text{on }\text{the }\text{left }\text{and }7\times \frac{9}{49}=\frac{9}{7} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
2+\frac{9}{7}=\frac{23}{7}: \\
7 \left(x^2-\frac{6 x}{7}+\frac{9}{49}\right)-6 \left(y^2-\frac{y}{6}+\underline{\text{ }}\right)=\fbox{$\frac{23}{7}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-1}{6}}{2}\right)^2=\frac{1}{144} \text{on }\text{the }\text{left }\text{and }\frac{-6}{144}=-\frac{1}{24} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{23}{7}-\frac{1}{24}=\frac{545}{168}: \\
7 \left(x^2-\frac{6 x}{7}+\frac{9}{49}\right)-6 \left(y^2-\frac{y}{6}+\frac{1}{144}\right)=\fbox{$\frac{545}{168}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{6 x}{7}+\frac{9}{49}=\left(x-\frac{3}{7}\right)^2: \\
7 \fbox{$\left(x-\frac{3}{7}\right)^2$}-6 \left(y^2-\frac{y}{6}+\frac{1}{144}\right)=\frac{545}{168} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{y}{6}+\frac{1}{144}=\left(y-\frac{1}{12}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 7 \left(x-\frac{3}{7}\right)^2-6 \fbox{$\left(y-\frac{1}{12}\right)^2$}=\frac{545}{168} \\
\end{array}
| khanacademy | amps |
Given the equation $-x^2+x+5 y^2+3 y+2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
5 y^2+3 y-x^2+x+2=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }2 \text{from }\text{both }\text{sides}: \\
5 y^2+3 y-x^2+x=-2 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-x^2+x+\underline{\text{ }}\right)+\left(5 y^2+3 y+\underline{\text{ }}\right)=\underline{\text{ }}-2 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-x^2+x+\underline{\text{ }}\right)=-\left(x^2-x+\underline{\text{ }}\right): \\
\fbox{$-\left(x^2-x+\underline{\text{ }}\right)$}+\left(5 y^2+3 y+\underline{\text{ }}\right)=\underline{\text{ }}-2 \\
\end{array}
Step 5:
\begin{array}{l}
\left(5 y^2+3 y+\underline{\text{ }}\right)=5 \left(y^2+\frac{3 y}{5}+\underline{\text{ }}\right): \\
-\left(x^2-x+\underline{\text{ }}\right)+\fbox{$5 \left(y^2+\frac{3 y}{5}+\underline{\text{ }}\right)$}=\underline{\text{ }}-2 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-1 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{-1}{4}=-\frac{1}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-2-\frac{1}{4}=-\frac{9}{4}: \\
-\left(x^2-x+\frac{1}{4}\right)+5 \left(y^2+\frac{3 y}{5}+\underline{\text{ }}\right)=\fbox{$-\frac{9}{4}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{3}{5}}{2}\right)^2=\frac{9}{100} \text{on }\text{the }\text{left }\text{and }5\times \frac{9}{100}=\frac{9}{20} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{9}{20}-\frac{9}{4}=-\frac{9}{5}: \\
-\left(x^2-x+\frac{1}{4}\right)+5 \left(y^2+\frac{3 y}{5}+\frac{9}{100}\right)=\fbox{$-\frac{9}{5}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-x+\frac{1}{4}=\left(x-\frac{1}{2}\right)^2: \\
-\fbox{$\left(x-\frac{1}{2}\right)^2$}+5 \left(y^2+\frac{3 y}{5}+\frac{9}{100}\right)=-\frac{9}{5} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{3 y}{5}+\frac{9}{100}=\left(y+\frac{3}{10}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -\left(x-\frac{1}{2}\right)^2+5 \fbox{$\left(y+\frac{3}{10}\right)^2$}=-\frac{9}{5} \\
\end{array}
| khanacademy | amps |
Given the equation $-4 x^2-2 x+8 y^2-7 y-9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
8 y^2-7 y-4 x^2-2 x-9=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }9 \text{to }\text{both }\text{sides}: \\
8 y^2-7 y-4 x^2-2 x=9 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-4 x^2-2 x+\underline{\text{ }}\right)+\left(8 y^2-7 y+\underline{\text{ }}\right)=\underline{\text{ }}+9 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-4 x^2-2 x+\underline{\text{ }}\right)=-4 \left(x^2+\frac{x}{2}+\underline{\text{ }}\right): \\
\fbox{$-4 \left(x^2+\frac{x}{2}+\underline{\text{ }}\right)$}+\left(8 y^2-7 y+\underline{\text{ }}\right)=\underline{\text{ }}+9 \\
\end{array}
Step 5:
\begin{array}{l}
\left(8 y^2-7 y+\underline{\text{ }}\right)=8 \left(y^2-\frac{7 y}{8}+\underline{\text{ }}\right): \\
-4 \left(x^2+\frac{x}{2}+\underline{\text{ }}\right)+\fbox{$8 \left(y^2-\frac{7 y}{8}+\underline{\text{ }}\right)$}=\underline{\text{ }}+9 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{2}}{2}\right)^2=\frac{1}{16} \text{on }\text{the }\text{left }\text{and }\frac{-4}{16}=-\frac{1}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
9-\frac{1}{4}=\frac{35}{4}: \\
-4 \left(x^2+\frac{x}{2}+\frac{1}{16}\right)+8 \left(y^2-\frac{7 y}{8}+\underline{\text{ }}\right)=\fbox{$\frac{35}{4}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-7}{8}}{2}\right)^2=\frac{49}{256} \text{on }\text{the }\text{left }\text{and }8\times \frac{49}{256}=\frac{49}{32} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{35}{4}+\frac{49}{32}=\frac{329}{32}: \\
-4 \left(x^2+\frac{x}{2}+\frac{1}{16}\right)+8 \left(y^2-\frac{7 y}{8}+\frac{49}{256}\right)=\fbox{$\frac{329}{32}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{x}{2}+\frac{1}{16}=\left(x+\frac{1}{4}\right)^2: \\
-4 \fbox{$\left(x+\frac{1}{4}\right)^2$}+8 \left(y^2-\frac{7 y}{8}+\frac{49}{256}\right)=\frac{329}{32} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{7 y}{8}+\frac{49}{256}=\left(y-\frac{7}{16}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -4 \left(x+\frac{1}{4}\right)^2+8 \fbox{$\left(y-\frac{7}{16}\right)^2$}=\frac{329}{32} \\
\end{array}
| khanacademy | amps |
Given the equation $-x^2+3 x-3 y^2-2 y+7=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-3 y^2-2 y-x^2+3 x+7=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }7 \text{from }\text{both }\text{sides}: \\
-3 y^2-2 y-x^2+3 x=-7 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-x^2+3 x+\underline{\text{ }}\right)+\left(-3 y^2-2 y+\underline{\text{ }}\right)=\underline{\text{ }}-7 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-x^2+3 x+\underline{\text{ }}\right)=-\left(x^2-3 x+\underline{\text{ }}\right): \\
\fbox{$-\left(x^2-3 x+\underline{\text{ }}\right)$}+\left(-3 y^2-2 y+\underline{\text{ }}\right)=\underline{\text{ }}-7 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-3 y^2-2 y+\underline{\text{ }}\right)=-3 \left(y^2+\frac{2 y}{3}+\underline{\text{ }}\right): \\
-\left(x^2-3 x+\underline{\text{ }}\right)+\fbox{$-3 \left(y^2+\frac{2 y}{3}+\underline{\text{ }}\right)$}=\underline{\text{ }}-7 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-1 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-3}{2}\right)^2=\frac{9}{4} \text{on }\text{the }\text{left }\text{and }-\frac{9}{4}=-\frac{9}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-7-\frac{9}{4}=-\frac{37}{4}: \\
-\left(x^2-3 x+\frac{9}{4}\right)-3 \left(y^2+\frac{2 y}{3}+\underline{\text{ }}\right)=\fbox{$-\frac{37}{4}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{2}{3}}{2}\right)^2=\frac{1}{9} \text{on }\text{the }\text{left }\text{and }\frac{-3}{9}=-\frac{1}{3} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{37}{4}-\frac{1}{3}=-\frac{115}{12}: \\
-\left(x^2-3 x+\frac{9}{4}\right)-3 \left(y^2+\frac{2 y}{3}+\frac{1}{9}\right)=\fbox{$-\frac{115}{12}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-3 x+\frac{9}{4}=\left(x-\frac{3}{2}\right)^2: \\
-\fbox{$\left(x-\frac{3}{2}\right)^2$}-3 \left(y^2+\frac{2 y}{3}+\frac{1}{9}\right)=-\frac{115}{12} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{2 y}{3}+\frac{1}{9}=\left(y+\frac{1}{3}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -\left(x-\frac{3}{2}\right)^2-3 \fbox{$\left(y+\frac{1}{3}\right)^2$}=-\frac{115}{12} \\
\end{array}
| khanacademy | amps |
Given the equation $-2 x^2-2 x-7 y^2-5 y+8=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-7 y^2-5 y-2 x^2-2 x+8=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }8 \text{from }\text{both }\text{sides}: \\
-7 y^2-5 y-2 x^2-2 x=-8 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-2 x^2-2 x+\underline{\text{ }}\right)+\left(-7 y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}-8 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-2 x^2-2 x+\underline{\text{ }}\right)=-2 \left(x^2+x+\underline{\text{ }}\right): \\
\fbox{$-2 \left(x^2+x+\underline{\text{ }}\right)$}+\left(-7 y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}-8 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-7 y^2-5 y+\underline{\text{ }}\right)=-7 \left(y^2+\frac{5 y}{7}+\underline{\text{ }}\right): \\
-2 \left(x^2+x+\underline{\text{ }}\right)+\fbox{$-7 \left(y^2+\frac{5 y}{7}+\underline{\text{ }}\right)$}=\underline{\text{ }}-8 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{-2}{4}=-\frac{1}{2} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-8-\frac{1}{2}=-\frac{17}{2}: \\
-2 \left(x^2+x+\frac{1}{4}\right)-7 \left(y^2+\frac{5 y}{7}+\underline{\text{ }}\right)=\fbox{$-\frac{17}{2}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{5}{7}}{2}\right)^2=\frac{25}{196} \text{on }\text{the }\text{left }\text{and }-7\times \frac{25}{196}=-\frac{25}{28} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{17}{2}-\frac{25}{28}=-\frac{263}{28}: \\
-2 \left(x^2+x+\frac{1}{4}\right)-7 \left(y^2+\frac{5 y}{7}+\frac{25}{196}\right)=\fbox{$-\frac{263}{28}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+x+\frac{1}{4}=\left(x+\frac{1}{2}\right)^2: \\
-2 \fbox{$\left(x+\frac{1}{2}\right)^2$}-7 \left(y^2+\frac{5 y}{7}+\frac{25}{196}\right)=-\frac{263}{28} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{5 y}{7}+\frac{25}{196}=\left(y+\frac{5}{14}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -2 \left(x+\frac{1}{2}\right)^2-7 \fbox{$\left(y+\frac{5}{14}\right)^2$}=-\frac{263}{28} \\
\end{array}
| khanacademy | amps |
Given the equation $-3 x^2-9 x-7 y^2-8 y+1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-7 y^2-8 y-3 x^2-9 x+1=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }1 \text{from }\text{both }\text{sides}: \\
-7 y^2-8 y-3 x^2-9 x=-1 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-3 x^2-9 x+\underline{\text{ }}\right)+\left(-7 y^2-8 y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-3 x^2-9 x+\underline{\text{ }}\right)=-3 \left(x^2+3 x+\underline{\text{ }}\right): \\
\fbox{$-3 \left(x^2+3 x+\underline{\text{ }}\right)$}+\left(-7 y^2-8 y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-7 y^2-8 y+\underline{\text{ }}\right)=-7 \left(y^2+\frac{8 y}{7}+\underline{\text{ }}\right): \\
-3 \left(x^2+3 x+\underline{\text{ }}\right)+\fbox{$-7 \left(y^2+\frac{8 y}{7}+\underline{\text{ }}\right)$}=\underline{\text{ }}-1 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{3}{2}\right)^2=\frac{9}{4} \text{on }\text{the }\text{left }\text{and }-3\times \frac{9}{4}=-\frac{27}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-1-\frac{27}{4}=-\frac{31}{4}: \\
-3 \left(x^2+3 x+\frac{9}{4}\right)-7 \left(y^2+\frac{8 y}{7}+\underline{\text{ }}\right)=\fbox{$-\frac{31}{4}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{8}{7}}{2}\right)^2=\frac{16}{49} \text{on }\text{the }\text{left }\text{and }-7\times \frac{16}{49}=-\frac{16}{7} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{31}{4}-\frac{16}{7}=-\frac{281}{28}: \\
-3 \left(x^2+3 x+\frac{9}{4}\right)-7 \left(y^2+\frac{8 y}{7}+\frac{16}{49}\right)=\fbox{$-\frac{281}{28}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+3 x+\frac{9}{4}=\left(x+\frac{3}{2}\right)^2: \\
-3 \fbox{$\left(x+\frac{3}{2}\right)^2$}-7 \left(y^2+\frac{8 y}{7}+\frac{16}{49}\right)=-\frac{281}{28} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{8 y}{7}+\frac{16}{49}=\left(y+\frac{4}{7}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -3 \left(x+\frac{3}{2}\right)^2-7 \fbox{$\left(y+\frac{4}{7}\right)^2$}=-\frac{281}{28} \\
\end{array}
| khanacademy | amps |
Given the equation $-x^2-8 x-3 y^2-3 y-10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-3 y^2-3 y-x^2-8 x-10=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }\text{10 }\text{to }\text{both }\text{sides}: \\
-3 y^2-3 y-x^2-8 x=10 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-x^2-8 x+\underline{\text{ }}\right)+\left(-3 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}+10 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-x^2-8 x+\underline{\text{ }}\right)=-\left(x^2+8 x+\underline{\text{ }}\right): \\
\fbox{$-\left(x^2+8 x+\underline{\text{ }}\right)$}+\left(-3 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}+10 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-3 y^2-3 y+\underline{\text{ }}\right)=-3 \left(y^2+y+\underline{\text{ }}\right): \\
-\left(x^2+8 x+\underline{\text{ }}\right)+\fbox{$-3 \left(y^2+y+\underline{\text{ }}\right)$}=\underline{\text{ }}+10 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-1 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{8}{2}\right)^2=16 \text{on }\text{the }\text{left }\text{and }-16=-16 \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
10-16=-6: \\
-\left(x^2+8 x+16\right)-3 \left(y^2+y+\underline{\text{ }}\right)=\fbox{$-6$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{-3}{4}=-\frac{3}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-6-\frac{3}{4}=-\frac{27}{4}: \\
-\left(x^2+8 x+16\right)-3 \left(y^2+y+\frac{1}{4}\right)=\fbox{$-\frac{27}{4}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+8 x+16=(x+4)^2: \\
-\fbox{$(x+4)^2$}-3 \left(y^2+y+\frac{1}{4}\right)=-\frac{27}{4} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+y+\frac{1}{4}=\left(y+\frac{1}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -(x+4)^2-3 \fbox{$\left(y+\frac{1}{2}\right)^2$}=-\frac{27}{4} \\
\end{array}
| khanacademy | amps |
Given the equation $-5 x^2+10 x-5 y^2-10 y-5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-5 y^2-10 y-5 x^2+10 x-5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }5 \text{to }\text{both }\text{sides}: \\
-5 y^2-10 y-5 x^2+10 x=5 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-5 x^2+10 x+\underline{\text{ }}\right)+\left(-5 y^2-10 y+\underline{\text{ }}\right)=\underline{\text{ }}+5 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-5 x^2+10 x+\underline{\text{ }}\right)=-5 \left(x^2-2 x+\underline{\text{ }}\right): \\
\fbox{$-5 \left(x^2-2 x+\underline{\text{ }}\right)$}+\left(-5 y^2-10 y+\underline{\text{ }}\right)=\underline{\text{ }}+5 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-5 y^2-10 y+\underline{\text{ }}\right)=-5 \left(y^2+2 y+\underline{\text{ }}\right): \\
-5 \left(x^2-2 x+\underline{\text{ }}\right)+\fbox{$-5 \left(y^2+2 y+\underline{\text{ }}\right)$}=\underline{\text{ }}+5 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-2}{2}\right)^2=1 \text{on }\text{the }\text{left }\text{and }-5\times 1=-5 \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
5-5=0: \\
-5 \left(x^2-2 x+1\right)-5 \left(y^2+2 y+\underline{\text{ }}\right)=\fbox{$0$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{2}{2}\right)^2=1 \text{on }\text{the }\text{left }\text{and }-5\times 1=-5 \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
x^2-2 x+1=(x-1)^2: \\
-5 \fbox{$(x-1)^2$}-5 \left(y^2+2 y+1\right)=-5 \\
\end{array}
Step 10:
\begin{array}{l}
y^2+2 y+1=(y+1)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -5 (x-1)^2-5 \fbox{$(y+1)^2$}=-5 \\
\end{array}
| khanacademy | amps |
Given the equation $-x^2-9 x-10 y^2-2 y+5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-10 y^2-2 y-x^2-9 x+5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }5 \text{from }\text{both }\text{sides}: \\
-10 y^2-2 y-x^2-9 x=-5 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-x^2-9 x+\underline{\text{ }}\right)+\left(-10 y^2-2 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-x^2-9 x+\underline{\text{ }}\right)=-\left(x^2+9 x+\underline{\text{ }}\right): \\
\fbox{$-\left(x^2+9 x+\underline{\text{ }}\right)$}+\left(-10 y^2-2 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-10 y^2-2 y+\underline{\text{ }}\right)=-10 \left(y^2+\frac{y}{5}+\underline{\text{ }}\right): \\
-\left(x^2+9 x+\underline{\text{ }}\right)+\fbox{$-10 \left(y^2+\frac{y}{5}+\underline{\text{ }}\right)$}=\underline{\text{ }}-5 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-1 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{9}{2}\right)^2=\frac{81}{4} \text{on }\text{the }\text{left }\text{and }-\frac{81}{4}=-\frac{81}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-5-\frac{81}{4}=-\frac{101}{4}: \\
-\left(x^2+9 x+\frac{81}{4}\right)-10 \left(y^2+\frac{y}{5}+\underline{\text{ }}\right)=\fbox{$-\frac{101}{4}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-10 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{5}}{2}\right)^2=\frac{1}{100} \text{on }\text{the }\text{left }\text{and }\frac{-10}{100}=-\frac{1}{10} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{101}{4}-\frac{1}{10}=-\frac{507}{20}: \\
-\left(x^2+9 x+\frac{81}{4}\right)-10 \left(y^2+\frac{y}{5}+\frac{1}{100}\right)=\fbox{$-\frac{507}{20}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+9 x+\frac{81}{4}=\left(x+\frac{9}{2}\right)^2: \\
-\fbox{$\left(x+\frac{9}{2}\right)^2$}-10 \left(y^2+\frac{y}{5}+\frac{1}{100}\right)=-\frac{507}{20} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{y}{5}+\frac{1}{100}=\left(y+\frac{1}{10}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -\left(x+\frac{9}{2}\right)^2-\text{10 }\fbox{$\left(y+\frac{1}{10}\right)^2$}=-\frac{507}{20} \\
\end{array}
| khanacademy | amps |
Given the equation $2 x^2+2 x-6 y^2+2 y-5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-6 y^2+2 y+2 x^2+2 x-5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }5 \text{to }\text{both }\text{sides}: \\
-6 y^2+2 y+2 x^2+2 x=5 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(2 x^2+2 x+\underline{\text{ }}\right)+\left(-6 y^2+2 y+\underline{\text{ }}\right)=\underline{\text{ }}+5 \\
\end{array}
Step 4:
\begin{array}{l}
\left(2 x^2+2 x+\underline{\text{ }}\right)=2 \left(x^2+x+\underline{\text{ }}\right): \\
\fbox{$2 \left(x^2+x+\underline{\text{ }}\right)$}+\left(-6 y^2+2 y+\underline{\text{ }}\right)=\underline{\text{ }}+5 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-6 y^2+2 y+\underline{\text{ }}\right)=-6 \left(y^2-\frac{y}{3}+\underline{\text{ }}\right): \\
2 \left(x^2+x+\underline{\text{ }}\right)+\fbox{$-6 \left(y^2-\frac{y}{3}+\underline{\text{ }}\right)$}=\underline{\text{ }}+5 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{2}{4}=\frac{1}{2} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
5+\frac{1}{2}=\frac{11}{2}: \\
2 \left(x^2+x+\frac{1}{4}\right)-6 \left(y^2-\frac{y}{3}+\underline{\text{ }}\right)=\fbox{$\frac{11}{2}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-1}{3}}{2}\right)^2=\frac{1}{36} \text{on }\text{the }\text{left }\text{and }\frac{-6}{36}=-\frac{1}{6} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{11}{2}-\frac{1}{6}=\frac{16}{3}: \\
2 \left(x^2+x+\frac{1}{4}\right)-6 \left(y^2-\frac{y}{3}+\frac{1}{36}\right)=\fbox{$\frac{16}{3}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+x+\frac{1}{4}=\left(x+\frac{1}{2}\right)^2: \\
2 \fbox{$\left(x+\frac{1}{2}\right)^2$}-6 \left(y^2-\frac{y}{3}+\frac{1}{36}\right)=\frac{16}{3} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{y}{3}+\frac{1}{36}=\left(y-\frac{1}{6}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 2 \left(x+\frac{1}{2}\right)^2-6 \fbox{$\left(y-\frac{1}{6}\right)^2$}=\frac{16}{3} \\
\end{array}
| khanacademy | amps |
Given the equation $2 x^2+3 x-5 y^2+y+2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-5 y^2+y+2 x^2+3 x+2=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }2 \text{from }\text{both }\text{sides}: \\
-5 y^2+y+2 x^2+3 x=-2 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(2 x^2+3 x+\underline{\text{ }}\right)+\left(-5 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}-2 \\
\end{array}
Step 4:
\begin{array}{l}
\left(2 x^2+3 x+\underline{\text{ }}\right)=2 \left(x^2+\frac{3 x}{2}+\underline{\text{ }}\right): \\
\fbox{$2 \left(x^2+\frac{3 x}{2}+\underline{\text{ }}\right)$}+\left(-5 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}-2 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-5 y^2+y+\underline{\text{ }}\right)=-5 \left(y^2-\frac{y}{5}+\underline{\text{ }}\right): \\
2 \left(x^2+\frac{3 x}{2}+\underline{\text{ }}\right)+\fbox{$-5 \left(y^2-\frac{y}{5}+\underline{\text{ }}\right)$}=\underline{\text{ }}-2 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{3}{2}}{2}\right)^2=\frac{9}{16} \text{on }\text{the }\text{left }\text{and }2\times \frac{9}{16}=\frac{9}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{9}{8}-2=-\frac{7}{8}: \\
2 \left(x^2+\frac{3 x}{2}+\frac{9}{16}\right)-5 \left(y^2-\frac{y}{5}+\underline{\text{ }}\right)=\fbox{$-\frac{7}{8}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-1}{5}}{2}\right)^2=\frac{1}{100} \text{on }\text{the }\text{left }\text{and }\frac{-5}{100}=-\frac{1}{20} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{7}{8}-\frac{1}{20}=-\frac{37}{40}: \\
2 \left(x^2+\frac{3 x}{2}+\frac{9}{16}\right)-5 \left(y^2-\frac{y}{5}+\frac{1}{100}\right)=\fbox{$-\frac{37}{40}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{3 x}{2}+\frac{9}{16}=\left(x+\frac{3}{4}\right)^2: \\
2 \fbox{$\left(x+\frac{3}{4}\right)^2$}-5 \left(y^2-\frac{y}{5}+\frac{1}{100}\right)=-\frac{37}{40} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{y}{5}+\frac{1}{100}=\left(y-\frac{1}{10}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 2 \left(x+\frac{3}{4}\right)^2-5 \fbox{$\left(y-\frac{1}{10}\right)^2$}=-\frac{37}{40} \\
\end{array}
| khanacademy | amps |
Given the equation $-10 x^2-7 x+5 y^2+10 y-8=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
5 y^2+10 y-10 x^2-7 x-8=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }8 \text{to }\text{both }\text{sides}: \\
5 y^2+10 y-10 x^2-7 x=8 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-10 x^2-7 x+\underline{\text{ }}\right)+\left(5 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}+8 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-10 x^2-7 x+\underline{\text{ }}\right)=-10 \left(x^2+\frac{7 x}{10}+\underline{\text{ }}\right): \\
\fbox{$-10 \left(x^2+\frac{7 x}{10}+\underline{\text{ }}\right)$}+\left(5 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}+8 \\
\end{array}
Step 5:
\begin{array}{l}
\left(5 y^2+10 y+\underline{\text{ }}\right)=5 \left(y^2+2 y+\underline{\text{ }}\right): \\
-10 \left(x^2+\frac{7 x}{10}+\underline{\text{ }}\right)+\fbox{$5 \left(y^2+2 y+\underline{\text{ }}\right)$}=\underline{\text{ }}+8 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-10 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{7}{10}}{2}\right)^2=\frac{49}{400} \text{on }\text{the }\text{left }\text{and }-10\times \frac{49}{400}=-\frac{49}{40} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
8-\frac{49}{40}=\frac{271}{40}: \\
-10 \left(x^2+\frac{7 x}{10}+\frac{49}{400}\right)+5 \left(y^2+2 y+\underline{\text{ }}\right)=\fbox{$\frac{271}{40}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{2}{2}\right)^2=1 \text{on }\text{the }\text{left }\text{and }5\times 1=5 \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{271}{40}+5=\frac{471}{40}: \\
-10 \left(x^2+\frac{7 x}{10}+\frac{49}{400}\right)+5 \left(y^2+2 y+1\right)=\fbox{$\frac{471}{40}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{7 x}{10}+\frac{49}{400}=\left(x+\frac{7}{20}\right)^2: \\
-10 \fbox{$\left(x+\frac{7}{20}\right)^2$}+5 \left(y^2+2 y+1\right)=\frac{471}{40} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+2 y+1=(y+1)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -10 \left(x+\frac{7}{20}\right)^2+5 \fbox{$(y+1)^2$}=\frac{471}{40} \\
\end{array}
| khanacademy | amps |
Given the equation $8 x^2+3 x-10 y^2+10 y+7=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-10 y^2+10 y+8 x^2+3 x+7=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }7 \text{from }\text{both }\text{sides}: \\
-10 y^2+10 y+8 x^2+3 x=-7 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(8 x^2+3 x+\underline{\text{ }}\right)+\left(-10 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}-7 \\
\end{array}
Step 4:
\begin{array}{l}
\left(8 x^2+3 x+\underline{\text{ }}\right)=8 \left(x^2+\frac{3 x}{8}+\underline{\text{ }}\right): \\
\fbox{$8 \left(x^2+\frac{3 x}{8}+\underline{\text{ }}\right)$}+\left(-10 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}-7 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-10 y^2+10 y+\underline{\text{ }}\right)=-10 \left(y^2-y+\underline{\text{ }}\right): \\
8 \left(x^2+\frac{3 x}{8}+\underline{\text{ }}\right)+\fbox{$-10 \left(y^2-y+\underline{\text{ }}\right)$}=\underline{\text{ }}-7 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{3}{8}}{2}\right)^2=\frac{9}{256} \text{on }\text{the }\text{left }\text{and }8\times \frac{9}{256}=\frac{9}{32} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{9}{32}-7=-\frac{215}{32}: \\
8 \left(x^2+\frac{3 x}{8}+\frac{9}{256}\right)-10 \left(y^2-y+\underline{\text{ }}\right)=\fbox{$-\frac{215}{32}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-10 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{-10}{4}=-\frac{5}{2} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{215}{32}-\frac{5}{2}=-\frac{295}{32}: \\
8 \left(x^2+\frac{3 x}{8}+\frac{9}{256}\right)-10 \left(y^2-y+\frac{1}{4}\right)=\fbox{$-\frac{295}{32}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{3 x}{8}+\frac{9}{256}=\left(x+\frac{3}{16}\right)^2: \\
8 \fbox{$\left(x+\frac{3}{16}\right)^2$}-10 \left(y^2-y+\frac{1}{4}\right)=-\frac{295}{32} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-y+\frac{1}{4}=\left(y-\frac{1}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 8 \left(x+\frac{3}{16}\right)^2-\text{10 }\fbox{$\left(y-\frac{1}{2}\right)^2$}=-\frac{295}{32} \\
\end{array}
| khanacademy | amps |
Given the equation $-8 x^2-5 x-7 y^2-y-1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-7 y^2-y-8 x^2-5 x-1=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }1 \text{to }\text{both }\text{sides}: \\
-7 y^2-y-8 x^2-5 x=1 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-8 x^2-5 x+\underline{\text{ }}\right)+\left(-7 y^2-y+\underline{\text{ }}\right)=\underline{\text{ }}+1 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-8 x^2-5 x+\underline{\text{ }}\right)=-8 \left(x^2+\frac{5 x}{8}+\underline{\text{ }}\right): \\
\fbox{$-8 \left(x^2+\frac{5 x}{8}+\underline{\text{ }}\right)$}+\left(-7 y^2-y+\underline{\text{ }}\right)=\underline{\text{ }}+1 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-7 y^2-y+\underline{\text{ }}\right)=-7 \left(y^2+\frac{y}{7}+\underline{\text{ }}\right): \\
-8 \left(x^2+\frac{5 x}{8}+\underline{\text{ }}\right)+\fbox{$-7 \left(y^2+\frac{y}{7}+\underline{\text{ }}\right)$}=\underline{\text{ }}+1 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{5}{8}}{2}\right)^2=\frac{25}{256} \text{on }\text{the }\text{left }\text{and }-8\times \frac{25}{256}=-\frac{25}{32} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
1-\frac{25}{32}=\frac{7}{32}: \\
-8 \left(x^2+\frac{5 x}{8}+\frac{25}{256}\right)-7 \left(y^2+\frac{y}{7}+\underline{\text{ }}\right)=\fbox{$\frac{7}{32}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{7}}{2}\right)^2=\frac{1}{196} \text{on }\text{the }\text{left }\text{and }\frac{-7}{196}=-\frac{1}{28} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{7}{32}-\frac{1}{28}=\frac{41}{224}: \\
-8 \left(x^2+\frac{5 x}{8}+\frac{25}{256}\right)-7 \left(y^2+\frac{y}{7}+\frac{1}{196}\right)=\fbox{$\frac{41}{224}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{5 x}{8}+\frac{25}{256}=\left(x+\frac{5}{16}\right)^2: \\
-8 \fbox{$\left(x+\frac{5}{16}\right)^2$}-7 \left(y^2+\frac{y}{7}+\frac{1}{196}\right)=\frac{41}{224} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{y}{7}+\frac{1}{196}=\left(y+\frac{1}{14}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -8 \left(x+\frac{5}{16}\right)^2-7 \fbox{$\left(y+\frac{1}{14}\right)^2$}=\frac{41}{224} \\
\end{array}
| khanacademy | amps |
Given the equation $-x^2+4 x+7 y^2-5 y+10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
7 y^2-5 y-x^2+4 x+10=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }\text{10 }\text{from }\text{both }\text{sides}: \\
7 y^2-5 y-x^2+4 x=-10 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-x^2+4 x+\underline{\text{ }}\right)+\left(7 y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}-10 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-x^2+4 x+\underline{\text{ }}\right)=-\left(x^2-4 x+\underline{\text{ }}\right): \\
\fbox{$-\left(x^2-4 x+\underline{\text{ }}\right)$}+\left(7 y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}-10 \\
\end{array}
Step 5:
\begin{array}{l}
\left(7 y^2-5 y+\underline{\text{ }}\right)=7 \left(y^2-\frac{5 y}{7}+\underline{\text{ }}\right): \\
-\left(x^2-4 x+\underline{\text{ }}\right)+\fbox{$7 \left(y^2-\frac{5 y}{7}+\underline{\text{ }}\right)$}=\underline{\text{ }}-10 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-1 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-4}{2}\right)^2=4 \text{on }\text{the }\text{left }\text{and }-4=-4 \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-10-4=-14: \\
-\left(x^2-4 x+4\right)+7 \left(y^2-\frac{5 y}{7}+\underline{\text{ }}\right)=\fbox{$-14$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-5}{7}}{2}\right)^2=\frac{25}{196} \text{on }\text{the }\text{left }\text{and }7\times \frac{25}{196}=\frac{25}{28} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{25}{28}-14=-\frac{367}{28}: \\
-\left(x^2-4 x+4\right)+7 \left(y^2-\frac{5 y}{7}+\frac{25}{196}\right)=\fbox{$-\frac{367}{28}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-4 x+4=(x-2)^2: \\
-\fbox{$(x-2)^2$}+7 \left(y^2-\frac{5 y}{7}+\frac{25}{196}\right)=-\frac{367}{28} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{5 y}{7}+\frac{25}{196}=\left(y-\frac{5}{14}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -(x-2)^2+7 \fbox{$\left(y-\frac{5}{14}\right)^2$}=-\frac{367}{28} \\
\end{array}
| khanacademy | amps |
Given the equation $9 x^2-10 x+5 y^2+3 y-6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
5 y^2+3 y+9 x^2-10 x-6=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }6 \text{to }\text{both }\text{sides}: \\
5 y^2+3 y+9 x^2-10 x=6 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(9 x^2-10 x+\underline{\text{ }}\right)+\left(5 y^2+3 y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\end{array}
Step 4:
\begin{array}{l}
\left(9 x^2-10 x+\underline{\text{ }}\right)=9 \left(x^2-\frac{10 x}{9}+\underline{\text{ }}\right): \\
\fbox{$9 \left(x^2-\frac{10 x}{9}+\underline{\text{ }}\right)$}+\left(5 y^2+3 y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\end{array}
Step 5:
\begin{array}{l}
\left(5 y^2+3 y+\underline{\text{ }}\right)=5 \left(y^2+\frac{3 y}{5}+\underline{\text{ }}\right): \\
9 \left(x^2-\frac{10 x}{9}+\underline{\text{ }}\right)+\fbox{$5 \left(y^2+\frac{3 y}{5}+\underline{\text{ }}\right)$}=\underline{\text{ }}+6 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-10}{9}}{2}\right)^2=\frac{25}{81} \text{on }\text{the }\text{left }\text{and }9\times \frac{25}{81}=\frac{25}{9} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
6+\frac{25}{9}=\frac{79}{9}: \\
9 \left(x^2-\frac{10 x}{9}+\frac{25}{81}\right)+5 \left(y^2+\frac{3 y}{5}+\underline{\text{ }}\right)=\fbox{$\frac{79}{9}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{3}{5}}{2}\right)^2=\frac{9}{100} \text{on }\text{the }\text{left }\text{and }5\times \frac{9}{100}=\frac{9}{20} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{79}{9}+\frac{9}{20}=\frac{1661}{180}: \\
9 \left(x^2-\frac{10 x}{9}+\frac{25}{81}\right)+5 \left(y^2+\frac{3 y}{5}+\frac{9}{100}\right)=\fbox{$\frac{1661}{180}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{10 x}{9}+\frac{25}{81}=\left(x-\frac{5}{9}\right)^2: \\
9 \fbox{$\left(x-\frac{5}{9}\right)^2$}+5 \left(y^2+\frac{3 y}{5}+\frac{9}{100}\right)=\frac{1661}{180} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{3 y}{5}+\frac{9}{100}=\left(y+\frac{3}{10}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 9 \left(x-\frac{5}{9}\right)^2+5 \fbox{$\left(y+\frac{3}{10}\right)^2$}=\frac{1661}{180} \\
\end{array}
| khanacademy | amps |
Given the equation $7 x^2+6 x-10 y^2-2 y+5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-10 y^2-2 y+7 x^2+6 x+5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }5 \text{from }\text{both }\text{sides}: \\
-10 y^2-2 y+7 x^2+6 x=-5 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(7 x^2+6 x+\underline{\text{ }}\right)+\left(-10 y^2-2 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\end{array}
Step 4:
\begin{array}{l}
\left(7 x^2+6 x+\underline{\text{ }}\right)=7 \left(x^2+\frac{6 x}{7}+\underline{\text{ }}\right): \\
\fbox{$7 \left(x^2+\frac{6 x}{7}+\underline{\text{ }}\right)$}+\left(-10 y^2-2 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-10 y^2-2 y+\underline{\text{ }}\right)=-10 \left(y^2+\frac{y}{5}+\underline{\text{ }}\right): \\
7 \left(x^2+\frac{6 x}{7}+\underline{\text{ }}\right)+\fbox{$-10 \left(y^2+\frac{y}{5}+\underline{\text{ }}\right)$}=\underline{\text{ }}-5 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{6}{7}}{2}\right)^2=\frac{9}{49} \text{on }\text{the }\text{left }\text{and }7\times \frac{9}{49}=\frac{9}{7} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{9}{7}-5=-\frac{26}{7}: \\
7 \left(x^2+\frac{6 x}{7}+\frac{9}{49}\right)-10 \left(y^2+\frac{y}{5}+\underline{\text{ }}\right)=\fbox{$-\frac{26}{7}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-10 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{5}}{2}\right)^2=\frac{1}{100} \text{on }\text{the }\text{left }\text{and }\frac{-10}{100}=-\frac{1}{10} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{26}{7}-\frac{1}{10}=-\frac{267}{70}: \\
7 \left(x^2+\frac{6 x}{7}+\frac{9}{49}\right)-10 \left(y^2+\frac{y}{5}+\frac{1}{100}\right)=\fbox{$-\frac{267}{70}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{6 x}{7}+\frac{9}{49}=\left(x+\frac{3}{7}\right)^2: \\
7 \fbox{$\left(x+\frac{3}{7}\right)^2$}-10 \left(y^2+\frac{y}{5}+\frac{1}{100}\right)=-\frac{267}{70} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{y}{5}+\frac{1}{100}=\left(y+\frac{1}{10}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 7 \left(x+\frac{3}{7}\right)^2-\text{10 }\fbox{$\left(y+\frac{1}{10}\right)^2$}=-\frac{267}{70} \\
\end{array}
| khanacademy | amps |
Given the equation $4 x^2-9 x-8 y^2-8 y-5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-8 y^2-8 y+4 x^2-9 x-5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }5 \text{to }\text{both }\text{sides}: \\
-8 y^2-8 y+4 x^2-9 x=5 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(4 x^2-9 x+\underline{\text{ }}\right)+\left(-8 y^2-8 y+\underline{\text{ }}\right)=\underline{\text{ }}+5 \\
\end{array}
Step 4:
\begin{array}{l}
\left(4 x^2-9 x+\underline{\text{ }}\right)=4 \left(x^2-\frac{9 x}{4}+\underline{\text{ }}\right): \\
\fbox{$4 \left(x^2-\frac{9 x}{4}+\underline{\text{ }}\right)$}+\left(-8 y^2-8 y+\underline{\text{ }}\right)=\underline{\text{ }}+5 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-8 y^2-8 y+\underline{\text{ }}\right)=-8 \left(y^2+y+\underline{\text{ }}\right): \\
4 \left(x^2-\frac{9 x}{4}+\underline{\text{ }}\right)+\fbox{$-8 \left(y^2+y+\underline{\text{ }}\right)$}=\underline{\text{ }}+5 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-9}{4}}{2}\right)^2=\frac{81}{64} \text{on }\text{the }\text{left }\text{and }4\times \frac{81}{64}=\frac{81}{16} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
5+\frac{81}{16}=\frac{161}{16}: \\
4 \left(x^2-\frac{9 x}{4}+\frac{81}{64}\right)-8 \left(y^2+y+\underline{\text{ }}\right)=\fbox{$\frac{161}{16}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{-8}{4}=-2 \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{161}{16}-2=\frac{129}{16}: \\
4 \left(x^2-\frac{9 x}{4}+\frac{81}{64}\right)-8 \left(y^2+y+\frac{1}{4}\right)=\fbox{$\frac{129}{16}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{9 x}{4}+\frac{81}{64}=\left(x-\frac{9}{8}\right)^2: \\
4 \fbox{$\left(x-\frac{9}{8}\right)^2$}-8 \left(y^2+y+\frac{1}{4}\right)=\frac{129}{16} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+y+\frac{1}{4}=\left(y+\frac{1}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 4 \left(x-\frac{9}{8}\right)^2-8 \fbox{$\left(y+\frac{1}{2}\right)^2$}=\frac{129}{16} \\
\end{array}
| khanacademy | amps |
Given the equation $4 x^2-10 x-4 y^2-3 y-7=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-4 y^2-3 y+4 x^2-10 x-7=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }7 \text{to }\text{both }\text{sides}: \\
-4 y^2-3 y+4 x^2-10 x=7 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(4 x^2-10 x+\underline{\text{ }}\right)+\left(-4 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}+7 \\
\end{array}
Step 4:
\begin{array}{l}
\left(4 x^2-10 x+\underline{\text{ }}\right)=4 \left(x^2-\frac{5 x}{2}+\underline{\text{ }}\right): \\
\fbox{$4 \left(x^2-\frac{5 x}{2}+\underline{\text{ }}\right)$}+\left(-4 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}+7 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-4 y^2-3 y+\underline{\text{ }}\right)=-4 \left(y^2+\frac{3 y}{4}+\underline{\text{ }}\right): \\
4 \left(x^2-\frac{5 x}{2}+\underline{\text{ }}\right)+\fbox{$-4 \left(y^2+\frac{3 y}{4}+\underline{\text{ }}\right)$}=\underline{\text{ }}+7 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-5}{2}}{2}\right)^2=\frac{25}{16} \text{on }\text{the }\text{left }\text{and }4\times \frac{25}{16}=\frac{25}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
7+\frac{25}{4}=\frac{53}{4}: \\
4 \left(x^2-\frac{5 x}{2}+\frac{25}{16}\right)-4 \left(y^2+\frac{3 y}{4}+\underline{\text{ }}\right)=\fbox{$\frac{53}{4}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{3}{4}}{2}\right)^2=\frac{9}{64} \text{on }\text{the }\text{left }\text{and }-4\times \frac{9}{64}=-\frac{9}{16} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{53}{4}-\frac{9}{16}=\frac{203}{16}: \\
4 \left(x^2-\frac{5 x}{2}+\frac{25}{16}\right)-4 \left(y^2+\frac{3 y}{4}+\frac{9}{64}\right)=\fbox{$\frac{203}{16}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{5 x}{2}+\frac{25}{16}=\left(x-\frac{5}{4}\right)^2: \\
4 \fbox{$\left(x-\frac{5}{4}\right)^2$}-4 \left(y^2+\frac{3 y}{4}+\frac{9}{64}\right)=\frac{203}{16} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{3 y}{4}+\frac{9}{64}=\left(y+\frac{3}{8}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 4 \left(x-\frac{5}{4}\right)^2-4 \fbox{$\left(y+\frac{3}{8}\right)^2$}=\frac{203}{16} \\
\end{array}
| khanacademy | amps |
Given the equation $4 x^2-5 x-4 y^2-7 y-1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-4 y^2-7 y+4 x^2-5 x-1=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }1 \text{to }\text{both }\text{sides}: \\
-4 y^2-7 y+4 x^2-5 x=1 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(4 x^2-5 x+\underline{\text{ }}\right)+\left(-4 y^2-7 y+\underline{\text{ }}\right)=\underline{\text{ }}+1 \\
\end{array}
Step 4:
\begin{array}{l}
\left(4 x^2-5 x+\underline{\text{ }}\right)=4 \left(x^2-\frac{5 x}{4}+\underline{\text{ }}\right): \\
\fbox{$4 \left(x^2-\frac{5 x}{4}+\underline{\text{ }}\right)$}+\left(-4 y^2-7 y+\underline{\text{ }}\right)=\underline{\text{ }}+1 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-4 y^2-7 y+\underline{\text{ }}\right)=-4 \left(y^2+\frac{7 y}{4}+\underline{\text{ }}\right): \\
4 \left(x^2-\frac{5 x}{4}+\underline{\text{ }}\right)+\fbox{$-4 \left(y^2+\frac{7 y}{4}+\underline{\text{ }}\right)$}=\underline{\text{ }}+1 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-5}{4}}{2}\right)^2=\frac{25}{64} \text{on }\text{the }\text{left }\text{and }4\times \frac{25}{64}=\frac{25}{16} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
1+\frac{25}{16}=\frac{41}{16}: \\
4 \left(x^2-\frac{5 x}{4}+\frac{25}{64}\right)-4 \left(y^2+\frac{7 y}{4}+\underline{\text{ }}\right)=\fbox{$\frac{41}{16}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{7}{4}}{2}\right)^2=\frac{49}{64} \text{on }\text{the }\text{left }\text{and }-4\times \frac{49}{64}=-\frac{49}{16} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{41}{16}-\frac{49}{16}=-\frac{1}{2}: \\
4 \left(x^2-\frac{5 x}{4}+\frac{25}{64}\right)-4 \left(y^2+\frac{7 y}{4}+\frac{49}{64}\right)=\fbox{$-\frac{1}{2}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{5 x}{4}+\frac{25}{64}=\left(x-\frac{5}{8}\right)^2: \\
4 \fbox{$\left(x-\frac{5}{8}\right)^2$}-4 \left(y^2+\frac{7 y}{4}+\frac{49}{64}\right)=-\frac{1}{2} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{7 y}{4}+\frac{49}{64}=\left(y+\frac{7}{8}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 4 \left(x-\frac{5}{8}\right)^2-4 \fbox{$\left(y+\frac{7}{8}\right)^2$}=-\frac{1}{2} \\
\end{array}
| khanacademy | amps |
Given the equation $4 x^2+10 x+5 y^2+3 y+2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
5 y^2+3 y+4 x^2+10 x+2=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }2 \text{from }\text{both }\text{sides}: \\
5 y^2+3 y+4 x^2+10 x=-2 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(4 x^2+10 x+\underline{\text{ }}\right)+\left(5 y^2+3 y+\underline{\text{ }}\right)=\underline{\text{ }}-2 \\
\end{array}
Step 4:
\begin{array}{l}
\left(4 x^2+10 x+\underline{\text{ }}\right)=4 \left(x^2+\frac{5 x}{2}+\underline{\text{ }}\right): \\
\fbox{$4 \left(x^2+\frac{5 x}{2}+\underline{\text{ }}\right)$}+\left(5 y^2+3 y+\underline{\text{ }}\right)=\underline{\text{ }}-2 \\
\end{array}
Step 5:
\begin{array}{l}
\left(5 y^2+3 y+\underline{\text{ }}\right)=5 \left(y^2+\frac{3 y}{5}+\underline{\text{ }}\right): \\
4 \left(x^2+\frac{5 x}{2}+\underline{\text{ }}\right)+\fbox{$5 \left(y^2+\frac{3 y}{5}+\underline{\text{ }}\right)$}=\underline{\text{ }}-2 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{5}{2}}{2}\right)^2=\frac{25}{16} \text{on }\text{the }\text{left }\text{and }4\times \frac{25}{16}=\frac{25}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{25}{4}-2=\frac{17}{4}: \\
4 \left(x^2+\frac{5 x}{2}+\frac{25}{16}\right)+5 \left(y^2+\frac{3 y}{5}+\underline{\text{ }}\right)=\fbox{$\frac{17}{4}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{3}{5}}{2}\right)^2=\frac{9}{100} \text{on }\text{the }\text{left }\text{and }5\times \frac{9}{100}=\frac{9}{20} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{17}{4}+\frac{9}{20}=\frac{47}{10}: \\
4 \left(x^2+\frac{5 x}{2}+\frac{25}{16}\right)+5 \left(y^2+\frac{3 y}{5}+\frac{9}{100}\right)=\fbox{$\frac{47}{10}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{5 x}{2}+\frac{25}{16}=\left(x+\frac{5}{4}\right)^2: \\
4 \fbox{$\left(x+\frac{5}{4}\right)^2$}+5 \left(y^2+\frac{3 y}{5}+\frac{9}{100}\right)=\frac{47}{10} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{3 y}{5}+\frac{9}{100}=\left(y+\frac{3}{10}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 4 \left(x+\frac{5}{4}\right)^2+5 \fbox{$\left(y+\frac{3}{10}\right)^2$}=\frac{47}{10} \\
\end{array}
| khanacademy | amps |
Given the equation $4 x^2-9 x-9 y^2-3 y=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-9 y^2-3 y+4 x^2-9 x=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(4 x^2-9 x+\underline{\text{ }}\right)+\left(-9 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}+0 \\
\end{array}
Step 3:
\begin{array}{l}
\left(4 x^2-9 x+\underline{\text{ }}\right)=4 \left(x^2-\frac{9 x}{4}+\underline{\text{ }}\right): \\
\fbox{$4 \left(x^2-\frac{9 x}{4}+\underline{\text{ }}\right)$}+\left(-9 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}+0 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-9 y^2-3 y+\underline{\text{ }}\right)=-9 \left(y^2+\frac{y}{3}+\underline{\text{ }}\right): \\
4 \left(x^2-\frac{9 x}{4}+\underline{\text{ }}\right)+\fbox{$-9 \left(y^2+\frac{y}{3}+\underline{\text{ }}\right)$}=\underline{\text{ }}+0 \\
\end{array}
Step 5:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-9}{4}}{2}\right)^2=\frac{81}{64} \text{on }\text{the }\text{left }\text{and }4\times \frac{81}{64}=\frac{81}{16} \text{on }\text{the }\text{right}: \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{3}}{2}\right)^2=\frac{1}{36} \text{on }\text{the }\text{left }\text{and }\frac{-9}{36}=-\frac{1}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{81}{16}-\frac{1}{4}=\frac{77}{16}: \\
4 \left(x^2-\frac{9 x}{4}+\frac{81}{64}\right)-9 \left(y^2+\frac{y}{3}+\frac{1}{36}\right)=\fbox{$\frac{77}{16}$} \\
\end{array}
Step 8:
\begin{array}{l}
x^2-\frac{9 x}{4}+\frac{81}{64}=\left(x-\frac{9}{8}\right)^2: \\
4 \fbox{$\left(x-\frac{9}{8}\right)^2$}-9 \left(y^2+\frac{y}{3}+\frac{1}{36}\right)=\frac{77}{16} \\
\end{array}
Step 9:
\begin{array}{l}
y^2+\frac{y}{3}+\frac{1}{36}=\left(y+\frac{1}{6}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 4 \left(x-\frac{9}{8}\right)^2-9 \fbox{$\left(y+\frac{1}{6}\right)^2$}=\frac{77}{16} \\
\end{array}
| khanacademy | amps |
Given the equation $-7 x^2+5 x+y^2+6 y+2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
y^2+6 y-7 x^2+5 x+2=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }2 \text{from }\text{both }\text{sides}: \\
y^2+6 y-7 x^2+5 x=-2 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-7 x^2+5 x+\underline{\text{ }}\right)+\left(y^2+6 y+\underline{\text{ }}\right)=\underline{\text{ }}-2 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-7 x^2+5 x+\underline{\text{ }}\right)=-7 \left(x^2-\frac{5 x}{7}+\underline{\text{ }}\right): \\
\fbox{$-7 \left(x^2-\frac{5 x}{7}+\underline{\text{ }}\right)$}+\left(y^2+6 y+\underline{\text{ }}\right)=\underline{\text{ }}-2 \\
\end{array}
Step 5:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-5}{7}}{2}\right)^2=\frac{25}{196} \text{on }\text{the }\text{left }\text{and }-7\times \frac{25}{196}=-\frac{25}{28} \text{on }\text{the }\text{right}: \\
\end{array}
Step 6:
\begin{array}{l}
-2-\frac{25}{28}=-\frac{81}{28}: \\
-7 \left(x^2-\frac{5 x}{7}+\frac{25}{196}\right)+\left(y^2+6 y+\underline{\text{ }}\right)=\fbox{$-\frac{81}{28}$} \\
\end{array}
Step 7:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it, }\text{then }\text{add }\text{it }\text{to }\text{both }\text{sides.} \\
\text{Add }\left(\frac{6}{2}\right)^2=9 \text{to }\text{both }\text{sides}: \\
\end{array}
Step 8:
\begin{array}{l}
9-\frac{81}{28}=\frac{171}{28}: \\
-7 \left(x^2-\frac{5 x}{7}+\frac{25}{196}\right)+\left(y^2+6 y+9\right)=\fbox{$\frac{171}{28}$} \\
\end{array}
Step 9:
\begin{array}{l}
x^2-\frac{5 x}{7}+\frac{25}{196}=\left(x-\frac{5}{14}\right)^2: \\
-7 \fbox{$\left(x-\frac{5}{14}\right)^2$}+\left(y^2+6 y+9\right)=\frac{171}{28} \\
\end{array}
Step 10:
\begin{array}{l}
y^2+6 y+9=(y+3)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -7 \left(x-\frac{5}{14}\right)^2+\fbox{$(y+3)^2$}=\frac{171}{28} \\
\end{array}
| khanacademy | amps |
Given the equation $-6 x^2-10 x+4 y^2+y+10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
4 y^2+y-6 x^2-10 x+10=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }\text{10 }\text{from }\text{both }\text{sides}: \\
4 y^2+y-6 x^2-10 x=-10 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-6 x^2-10 x+\underline{\text{ }}\right)+\left(4 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}-10 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-6 x^2-10 x+\underline{\text{ }}\right)=-6 \left(x^2+\frac{5 x}{3}+\underline{\text{ }}\right): \\
\fbox{$-6 \left(x^2+\frac{5 x}{3}+\underline{\text{ }}\right)$}+\left(4 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}-10 \\
\end{array}
Step 5:
\begin{array}{l}
\left(4 y^2+y+\underline{\text{ }}\right)=4 \left(y^2+\frac{y}{4}+\underline{\text{ }}\right): \\
-6 \left(x^2+\frac{5 x}{3}+\underline{\text{ }}\right)+\fbox{$4 \left(y^2+\frac{y}{4}+\underline{\text{ }}\right)$}=\underline{\text{ }}-10 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{5}{3}}{2}\right)^2=\frac{25}{36} \text{on }\text{the }\text{left }\text{and }-6\times \frac{25}{36}=-\frac{25}{6} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-10-\frac{25}{6}=-\frac{85}{6}: \\
-6 \left(x^2+\frac{5 x}{3}+\frac{25}{36}\right)+4 \left(y^2+\frac{y}{4}+\underline{\text{ }}\right)=\fbox{$-\frac{85}{6}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{4}}{2}\right)^2=\frac{1}{64} \text{on }\text{the }\text{left }\text{and }\frac{4}{64}=\frac{1}{16} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{1}{16}-\frac{85}{6}=-\frac{677}{48}: \\
-6 \left(x^2+\frac{5 x}{3}+\frac{25}{36}\right)+4 \left(y^2+\frac{y}{4}+\frac{1}{64}\right)=\fbox{$-\frac{677}{48}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{5 x}{3}+\frac{25}{36}=\left(x+\frac{5}{6}\right)^2: \\
-6 \fbox{$\left(x+\frac{5}{6}\right)^2$}+4 \left(y^2+\frac{y}{4}+\frac{1}{64}\right)=-\frac{677}{48} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{y}{4}+\frac{1}{64}=\left(y+\frac{1}{8}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -6 \left(x+\frac{5}{6}\right)^2+4 \fbox{$\left(y+\frac{1}{8}\right)^2$}=-\frac{677}{48} \\
\end{array}
| khanacademy | amps |
Given the equation $-8 x^2+6 x+10 y^2+8 y-3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
10 y^2+8 y-8 x^2+6 x-3=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }3 \text{to }\text{both }\text{sides}: \\
10 y^2+8 y-8 x^2+6 x=3 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-8 x^2+6 x+\underline{\text{ }}\right)+\left(10 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}+3 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-8 x^2+6 x+\underline{\text{ }}\right)=-8 \left(x^2-\frac{3 x}{4}+\underline{\text{ }}\right): \\
\fbox{$-8 \left(x^2-\frac{3 x}{4}+\underline{\text{ }}\right)$}+\left(10 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}+3 \\
\end{array}
Step 5:
\begin{array}{l}
\left(10 y^2+8 y+\underline{\text{ }}\right)=10 \left(y^2+\frac{4 y}{5}+\underline{\text{ }}\right): \\
-8 \left(x^2-\frac{3 x}{4}+\underline{\text{ }}\right)+\fbox{$10 \left(y^2+\frac{4 y}{5}+\underline{\text{ }}\right)$}=\underline{\text{ }}+3 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-3}{4}}{2}\right)^2=\frac{9}{64} \text{on }\text{the }\text{left }\text{and }-8\times \frac{9}{64}=-\frac{9}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
3-\frac{9}{8}=\frac{15}{8}: \\
-8 \left(x^2-\frac{3 x}{4}+\frac{9}{64}\right)+10 \left(y^2+\frac{4 y}{5}+\underline{\text{ }}\right)=\fbox{$\frac{15}{8}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }\text{10 }\text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{4}{5}}{2}\right)^2=\frac{4}{25} \text{on }\text{the }\text{left }\text{and }10\times \frac{4}{25}=\frac{8}{5} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{15}{8}+\frac{8}{5}=\frac{139}{40}: \\
-8 \left(x^2-\frac{3 x}{4}+\frac{9}{64}\right)+10 \left(y^2+\frac{4 y}{5}+\frac{4}{25}\right)=\fbox{$\frac{139}{40}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{3 x}{4}+\frac{9}{64}=\left(x-\frac{3}{8}\right)^2: \\
-8 \fbox{$\left(x-\frac{3}{8}\right)^2$}+10 \left(y^2+\frac{4 y}{5}+\frac{4}{25}\right)=\frac{139}{40} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{4 y}{5}+\frac{4}{25}=\left(y+\frac{2}{5}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -8 \left(x-\frac{3}{8}\right)^2+\text{10 }\fbox{$\left(y+\frac{2}{5}\right)^2$}=\frac{139}{40} \\
\end{array}
| khanacademy | amps |
Given the equation $-9 x^2-8 x-9 y^2-5 y+1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-9 y^2-5 y-9 x^2-8 x+1=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }1 \text{from }\text{both }\text{sides}: \\
-9 y^2-5 y-9 x^2-8 x=-1 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-9 x^2-8 x+\underline{\text{ }}\right)+\left(-9 y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-9 x^2-8 x+\underline{\text{ }}\right)=-9 \left(x^2+\frac{8 x}{9}+\underline{\text{ }}\right): \\
\fbox{$-9 \left(x^2+\frac{8 x}{9}+\underline{\text{ }}\right)$}+\left(-9 y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-9 y^2-5 y+\underline{\text{ }}\right)=-9 \left(y^2+\frac{5 y}{9}+\underline{\text{ }}\right): \\
-9 \left(x^2+\frac{8 x}{9}+\underline{\text{ }}\right)+\fbox{$-9 \left(y^2+\frac{5 y}{9}+\underline{\text{ }}\right)$}=\underline{\text{ }}-1 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{8}{9}}{2}\right)^2=\frac{16}{81} \text{on }\text{the }\text{left }\text{and }-9\times \frac{16}{81}=-\frac{16}{9} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-1-\frac{16}{9}=-\frac{25}{9}: \\
-9 \left(x^2+\frac{8 x}{9}+\frac{16}{81}\right)-9 \left(y^2+\frac{5 y}{9}+\underline{\text{ }}\right)=\fbox{$-\frac{25}{9}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{5}{9}}{2}\right)^2=\frac{25}{324} \text{on }\text{the }\text{left }\text{and }-9\times \frac{25}{324}=-\frac{25}{36} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{25}{9}-\frac{25}{36}=-\frac{125}{36}: \\
-9 \left(x^2+\frac{8 x}{9}+\frac{16}{81}\right)-9 \left(y^2+\frac{5 y}{9}+\frac{25}{324}\right)=\fbox{$-\frac{125}{36}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{8 x}{9}+\frac{16}{81}=\left(x+\frac{4}{9}\right)^2: \\
-9 \fbox{$\left(x+\frac{4}{9}\right)^2$}-9 \left(y^2+\frac{5 y}{9}+\frac{25}{324}\right)=-\frac{125}{36} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{5 y}{9}+\frac{25}{324}=\left(y+\frac{5}{18}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -9 \left(x+\frac{4}{9}\right)^2-9 \fbox{$\left(y+\frac{5}{18}\right)^2$}=-\frac{125}{36} \\
\end{array}
| khanacademy | amps |
Given the equation $-3 x^2+5 x-y^2+6 y-2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-y^2+6 y-3 x^2+5 x-2=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }2 \text{to }\text{both }\text{sides}: \\
-y^2+6 y-3 x^2+5 x=2 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-3 x^2+5 x+\underline{\text{ }}\right)+\left(-y^2+6 y+\underline{\text{ }}\right)=\underline{\text{ }}+2 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-3 x^2+5 x+\underline{\text{ }}\right)=-3 \left(x^2-\frac{5 x}{3}+\underline{\text{ }}\right): \\
\fbox{$-3 \left(x^2-\frac{5 x}{3}+\underline{\text{ }}\right)$}+\left(-y^2+6 y+\underline{\text{ }}\right)=\underline{\text{ }}+2 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-y^2+6 y+\underline{\text{ }}\right)=-\left(y^2-6 y+\underline{\text{ }}\right): \\
-3 \left(x^2-\frac{5 x}{3}+\underline{\text{ }}\right)+\fbox{$-\left(y^2-6 y+\underline{\text{ }}\right)$}=\underline{\text{ }}+2 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-5}{3}}{2}\right)^2=\frac{25}{36} \text{on }\text{the }\text{left }\text{and }-3\times \frac{25}{36}=-\frac{25}{12} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
2-\frac{25}{12}=-\frac{1}{12}: \\
-3 \left(x^2-\frac{5 x}{3}+\frac{25}{36}\right)-\left(y^2-6 y+\underline{\text{ }}\right)=\fbox{$-\frac{1}{12}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-1 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-6}{2}\right)^2=9 \text{on }\text{the }\text{left }\text{and }-9=-9 \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{1}{12}-9=-\frac{109}{12}: \\
-3 \left(x^2-\frac{5 x}{3}+\frac{25}{36}\right)-\left(y^2-6 y+9\right)=\fbox{$-\frac{109}{12}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{5 x}{3}+\frac{25}{36}=\left(x-\frac{5}{6}\right)^2: \\
-3 \fbox{$\left(x-\frac{5}{6}\right)^2$}-\left(y^2-6 y+9\right)=-\frac{109}{12} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-6 y+9=(y-3)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -3 \left(x-\frac{5}{6}\right)^2-\fbox{$(y-3)^2$}=-\frac{109}{12} \\
\end{array}
| khanacademy | amps |
Given the equation $2 x^2-5 x+4 y^2-10 y-10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
4 y^2-10 y+2 x^2-5 x-10=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }\text{10 }\text{to }\text{both }\text{sides}: \\
4 y^2-10 y+2 x^2-5 x=10 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(2 x^2-5 x+\underline{\text{ }}\right)+\left(4 y^2-10 y+\underline{\text{ }}\right)=\underline{\text{ }}+10 \\
\end{array}
Step 4:
\begin{array}{l}
\left(2 x^2-5 x+\underline{\text{ }}\right)=2 \left(x^2-\frac{5 x}{2}+\underline{\text{ }}\right): \\
\fbox{$2 \left(x^2-\frac{5 x}{2}+\underline{\text{ }}\right)$}+\left(4 y^2-10 y+\underline{\text{ }}\right)=\underline{\text{ }}+10 \\
\end{array}
Step 5:
\begin{array}{l}
\left(4 y^2-10 y+\underline{\text{ }}\right)=4 \left(y^2-\frac{5 y}{2}+\underline{\text{ }}\right): \\
2 \left(x^2-\frac{5 x}{2}+\underline{\text{ }}\right)+\fbox{$4 \left(y^2-\frac{5 y}{2}+\underline{\text{ }}\right)$}=\underline{\text{ }}+10 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-5}{2}}{2}\right)^2=\frac{25}{16} \text{on }\text{the }\text{left }\text{and }2\times \frac{25}{16}=\frac{25}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
10+\frac{25}{8}=\frac{105}{8}: \\
2 \left(x^2-\frac{5 x}{2}+\frac{25}{16}\right)+4 \left(y^2-\frac{5 y}{2}+\underline{\text{ }}\right)=\fbox{$\frac{105}{8}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-5}{2}}{2}\right)^2=\frac{25}{16} \text{on }\text{the }\text{left }\text{and }4\times \frac{25}{16}=\frac{25}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{105}{8}+\frac{25}{4}=\frac{155}{8}: \\
2 \left(x^2-\frac{5 x}{2}+\frac{25}{16}\right)+4 \left(y^2-\frac{5 y}{2}+\frac{25}{16}\right)=\fbox{$\frac{155}{8}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{5 x}{2}+\frac{25}{16}=\left(x-\frac{5}{4}\right)^2: \\
2 \fbox{$\left(x-\frac{5}{4}\right)^2$}+4 \left(y^2-\frac{5 y}{2}+\frac{25}{16}\right)=\frac{155}{8} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{5 y}{2}+\frac{25}{16}=\left(y-\frac{5}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 2 \left(x-\frac{5}{4}\right)^2+4 \fbox{$\left(y-\frac{5}{4}\right)^2$}=\frac{155}{8} \\
\end{array}
| khanacademy | amps |
Given the equation $-10 x^2-3 x+6 y^2+10 y+8=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
6 y^2+10 y-10 x^2-3 x+8=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }8 \text{from }\text{both }\text{sides}: \\
6 y^2+10 y-10 x^2-3 x=-8 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-10 x^2-3 x+\underline{\text{ }}\right)+\left(6 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}-8 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-10 x^2-3 x+\underline{\text{ }}\right)=-10 \left(x^2+\frac{3 x}{10}+\underline{\text{ }}\right): \\
\fbox{$-10 \left(x^2+\frac{3 x}{10}+\underline{\text{ }}\right)$}+\left(6 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}-8 \\
\end{array}
Step 5:
\begin{array}{l}
\left(6 y^2+10 y+\underline{\text{ }}\right)=6 \left(y^2+\frac{5 y}{3}+\underline{\text{ }}\right): \\
-10 \left(x^2+\frac{3 x}{10}+\underline{\text{ }}\right)+\fbox{$6 \left(y^2+\frac{5 y}{3}+\underline{\text{ }}\right)$}=\underline{\text{ }}-8 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-10 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{3}{10}}{2}\right)^2=\frac{9}{400} \text{on }\text{the }\text{left }\text{and }-10\times \frac{9}{400}=-\frac{9}{40} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-8-\frac{9}{40}=-\frac{329}{40}: \\
-10 \left(x^2+\frac{3 x}{10}+\frac{9}{400}\right)+6 \left(y^2+\frac{5 y}{3}+\underline{\text{ }}\right)=\fbox{$-\frac{329}{40}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{5}{3}}{2}\right)^2=\frac{25}{36} \text{on }\text{the }\text{left }\text{and }6\times \frac{25}{36}=\frac{25}{6} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{25}{6}-\frac{329}{40}=-\frac{487}{120}: \\
-10 \left(x^2+\frac{3 x}{10}+\frac{9}{400}\right)+6 \left(y^2+\frac{5 y}{3}+\frac{25}{36}\right)=\fbox{$-\frac{487}{120}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{3 x}{10}+\frac{9}{400}=\left(x+\frac{3}{20}\right)^2: \\
-10 \fbox{$\left(x+\frac{3}{20}\right)^2$}+6 \left(y^2+\frac{5 y}{3}+\frac{25}{36}\right)=-\frac{487}{120} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{5 y}{3}+\frac{25}{36}=\left(y+\frac{5}{6}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -10 \left(x+\frac{3}{20}\right)^2+6 \fbox{$\left(y+\frac{5}{6}\right)^2$}=-\frac{487}{120} \\
\end{array}
| khanacademy | amps |
Given the equation $7 x^2+6 x-8 y^2+5 y+7=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-8 y^2+5 y+7 x^2+6 x+7=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }7 \text{from }\text{both }\text{sides}: \\
-8 y^2+5 y+7 x^2+6 x=-7 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(7 x^2+6 x+\underline{\text{ }}\right)+\left(-8 y^2+5 y+\underline{\text{ }}\right)=\underline{\text{ }}-7 \\
\end{array}
Step 4:
\begin{array}{l}
\left(7 x^2+6 x+\underline{\text{ }}\right)=7 \left(x^2+\frac{6 x}{7}+\underline{\text{ }}\right): \\
\fbox{$7 \left(x^2+\frac{6 x}{7}+\underline{\text{ }}\right)$}+\left(-8 y^2+5 y+\underline{\text{ }}\right)=\underline{\text{ }}-7 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-8 y^2+5 y+\underline{\text{ }}\right)=-8 \left(y^2-\frac{5 y}{8}+\underline{\text{ }}\right): \\
7 \left(x^2+\frac{6 x}{7}+\underline{\text{ }}\right)+\fbox{$-8 \left(y^2-\frac{5 y}{8}+\underline{\text{ }}\right)$}=\underline{\text{ }}-7 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{6}{7}}{2}\right)^2=\frac{9}{49} \text{on }\text{the }\text{left }\text{and }7\times \frac{9}{49}=\frac{9}{7} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{9}{7}-7=-\frac{40}{7}: \\
7 \left(x^2+\frac{6 x}{7}+\frac{9}{49}\right)-8 \left(y^2-\frac{5 y}{8}+\underline{\text{ }}\right)=\fbox{$-\frac{40}{7}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-5}{8}}{2}\right)^2=\frac{25}{256} \text{on }\text{the }\text{left }\text{and }-8\times \frac{25}{256}=-\frac{25}{32} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{40}{7}-\frac{25}{32}=-\frac{1455}{224}: \\
7 \left(x^2+\frac{6 x}{7}+\frac{9}{49}\right)-8 \left(y^2-\frac{5 y}{8}+\frac{25}{256}\right)=\fbox{$-\frac{1455}{224}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{6 x}{7}+\frac{9}{49}=\left(x+\frac{3}{7}\right)^2: \\
7 \fbox{$\left(x+\frac{3}{7}\right)^2$}-8 \left(y^2-\frac{5 y}{8}+\frac{25}{256}\right)=-\frac{1455}{224} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{5 y}{8}+\frac{25}{256}=\left(y-\frac{5}{16}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 7 \left(x+\frac{3}{7}\right)^2-8 \fbox{$\left(y-\frac{5}{16}\right)^2$}=-\frac{1455}{224} \\
\end{array}
| khanacademy | amps |
Given the equation $-3 x^2+3 x-9 y^2-5 y-6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-9 y^2-5 y-3 x^2+3 x-6=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }6 \text{to }\text{both }\text{sides}: \\
-9 y^2-5 y-3 x^2+3 x=6 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-3 x^2+3 x+\underline{\text{ }}\right)+\left(-9 y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-3 x^2+3 x+\underline{\text{ }}\right)=-3 \left(x^2-x+\underline{\text{ }}\right): \\
\fbox{$-3 \left(x^2-x+\underline{\text{ }}\right)$}+\left(-9 y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-9 y^2-5 y+\underline{\text{ }}\right)=-9 \left(y^2+\frac{5 y}{9}+\underline{\text{ }}\right): \\
-3 \left(x^2-x+\underline{\text{ }}\right)+\fbox{$-9 \left(y^2+\frac{5 y}{9}+\underline{\text{ }}\right)$}=\underline{\text{ }}+6 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{-3}{4}=-\frac{3}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
6-\frac{3}{4}=\frac{21}{4}: \\
-3 \left(x^2-x+\frac{1}{4}\right)-9 \left(y^2+\frac{5 y}{9}+\underline{\text{ }}\right)=\fbox{$\frac{21}{4}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{5}{9}}{2}\right)^2=\frac{25}{324} \text{on }\text{the }\text{left }\text{and }-9\times \frac{25}{324}=-\frac{25}{36} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{21}{4}-\frac{25}{36}=\frac{41}{9}: \\
-3 \left(x^2-x+\frac{1}{4}\right)-9 \left(y^2+\frac{5 y}{9}+\frac{25}{324}\right)=\fbox{$\frac{41}{9}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-x+\frac{1}{4}=\left(x-\frac{1}{2}\right)^2: \\
-3 \fbox{$\left(x-\frac{1}{2}\right)^2$}-9 \left(y^2+\frac{5 y}{9}+\frac{25}{324}\right)=\frac{41}{9} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{5 y}{9}+\frac{25}{324}=\left(y+\frac{5}{18}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -3 \left(x-\frac{1}{2}\right)^2-9 \fbox{$\left(y+\frac{5}{18}\right)^2$}=\frac{41}{9} \\
\end{array}
| khanacademy | amps |
Given the equation $-9 x^2-10 x+7 y^2-4 y+1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
7 y^2-4 y-9 x^2-10 x+1=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }1 \text{from }\text{both }\text{sides}: \\
7 y^2-4 y-9 x^2-10 x=-1 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-9 x^2-10 x+\underline{\text{ }}\right)+\left(7 y^2-4 y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-9 x^2-10 x+\underline{\text{ }}\right)=-9 \left(x^2+\frac{10 x}{9}+\underline{\text{ }}\right): \\
\fbox{$-9 \left(x^2+\frac{10 x}{9}+\underline{\text{ }}\right)$}+\left(7 y^2-4 y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\end{array}
Step 5:
\begin{array}{l}
\left(7 y^2-4 y+\underline{\text{ }}\right)=7 \left(y^2-\frac{4 y}{7}+\underline{\text{ }}\right): \\
-9 \left(x^2+\frac{10 x}{9}+\underline{\text{ }}\right)+\fbox{$7 \left(y^2-\frac{4 y}{7}+\underline{\text{ }}\right)$}=\underline{\text{ }}-1 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{10}{9}}{2}\right)^2=\frac{25}{81} \text{on }\text{the }\text{left }\text{and }-9\times \frac{25}{81}=-\frac{25}{9} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-1-\frac{25}{9}=-\frac{34}{9}: \\
-9 \left(x^2+\frac{10 x}{9}+\frac{25}{81}\right)+7 \left(y^2-\frac{4 y}{7}+\underline{\text{ }}\right)=\fbox{$-\frac{34}{9}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-4}{7}}{2}\right)^2=\frac{4}{49} \text{on }\text{the }\text{left }\text{and }7\times \frac{4}{49}=\frac{4}{7} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{4}{7}-\frac{34}{9}=-\frac{202}{63}: \\
-9 \left(x^2+\frac{10 x}{9}+\frac{25}{81}\right)+7 \left(y^2-\frac{4 y}{7}+\frac{4}{49}\right)=\fbox{$-\frac{202}{63}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{10 x}{9}+\frac{25}{81}=\left(x+\frac{5}{9}\right)^2: \\
-9 \fbox{$\left(x+\frac{5}{9}\right)^2$}+7 \left(y^2-\frac{4 y}{7}+\frac{4}{49}\right)=-\frac{202}{63} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{4 y}{7}+\frac{4}{49}=\left(y-\frac{2}{7}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -9 \left(x+\frac{5}{9}\right)^2+7 \fbox{$\left(y-\frac{2}{7}\right)^2$}=-\frac{202}{63} \\
\end{array}
| khanacademy | amps |
Given the equation $-4 x^2+5 x+8 y^2-4 y+7=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
8 y^2-4 y-4 x^2+5 x+7=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }7 \text{from }\text{both }\text{sides}: \\
8 y^2-4 y-4 x^2+5 x=-7 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-4 x^2+5 x+\underline{\text{ }}\right)+\left(8 y^2-4 y+\underline{\text{ }}\right)=\underline{\text{ }}-7 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-4 x^2+5 x+\underline{\text{ }}\right)=-4 \left(x^2-\frac{5 x}{4}+\underline{\text{ }}\right): \\
\fbox{$-4 \left(x^2-\frac{5 x}{4}+\underline{\text{ }}\right)$}+\left(8 y^2-4 y+\underline{\text{ }}\right)=\underline{\text{ }}-7 \\
\end{array}
Step 5:
\begin{array}{l}
\left(8 y^2-4 y+\underline{\text{ }}\right)=8 \left(y^2-\frac{y}{2}+\underline{\text{ }}\right): \\
-4 \left(x^2-\frac{5 x}{4}+\underline{\text{ }}\right)+\fbox{$8 \left(y^2-\frac{y}{2}+\underline{\text{ }}\right)$}=\underline{\text{ }}-7 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-5}{4}}{2}\right)^2=\frac{25}{64} \text{on }\text{the }\text{left }\text{and }-4\times \frac{25}{64}=-\frac{25}{16} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-7-\frac{25}{16}=-\frac{137}{16}: \\
-4 \left(x^2-\frac{5 x}{4}+\frac{25}{64}\right)+8 \left(y^2-\frac{y}{2}+\underline{\text{ }}\right)=\fbox{$-\frac{137}{16}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-1}{2}}{2}\right)^2=\frac{1}{16} \text{on }\text{the }\text{left }\text{and }\frac{8}{16}=\frac{1}{2} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{1}{2}-\frac{137}{16}=-\frac{129}{16}: \\
-4 \left(x^2-\frac{5 x}{4}+\frac{25}{64}\right)+8 \left(y^2-\frac{y}{2}+\frac{1}{16}\right)=\fbox{$-\frac{129}{16}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{5 x}{4}+\frac{25}{64}=\left(x-\frac{5}{8}\right)^2: \\
-4 \fbox{$\left(x-\frac{5}{8}\right)^2$}+8 \left(y^2-\frac{y}{2}+\frac{1}{16}\right)=-\frac{129}{16} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{y}{2}+\frac{1}{16}=\left(y-\frac{1}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -4 \left(x-\frac{5}{8}\right)^2+8 \fbox{$\left(y-\frac{1}{4}\right)^2$}=-\frac{129}{16} \\
\end{array}
| khanacademy | amps |
Given the equation $6 x^2-3 x+2 y^2-9 y+5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
2 y^2-9 y+6 x^2-3 x+5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }5 \text{from }\text{both }\text{sides}: \\
2 y^2-9 y+6 x^2-3 x=-5 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(6 x^2-3 x+\underline{\text{ }}\right)+\left(2 y^2-9 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\end{array}
Step 4:
\begin{array}{l}
\left(6 x^2-3 x+\underline{\text{ }}\right)=6 \left(x^2-\frac{x}{2}+\underline{\text{ }}\right): \\
\fbox{$6 \left(x^2-\frac{x}{2}+\underline{\text{ }}\right)$}+\left(2 y^2-9 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\end{array}
Step 5:
\begin{array}{l}
\left(2 y^2-9 y+\underline{\text{ }}\right)=2 \left(y^2-\frac{9 y}{2}+\underline{\text{ }}\right): \\
6 \left(x^2-\frac{x}{2}+\underline{\text{ }}\right)+\fbox{$2 \left(y^2-\frac{9 y}{2}+\underline{\text{ }}\right)$}=\underline{\text{ }}-5 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-1}{2}}{2}\right)^2=\frac{1}{16} \text{on }\text{the }\text{left }\text{and }\frac{6}{16}=\frac{3}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{3}{8}-5=-\frac{37}{8}: \\
6 \left(x^2-\frac{x}{2}+\frac{1}{16}\right)+2 \left(y^2-\frac{9 y}{2}+\underline{\text{ }}\right)=\fbox{$-\frac{37}{8}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-9}{2}}{2}\right)^2=\frac{81}{16} \text{on }\text{the }\text{left }\text{and }2\times \frac{81}{16}=\frac{81}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{81}{8}-\frac{37}{8}=\frac{11}{2}: \\
6 \left(x^2-\frac{x}{2}+\frac{1}{16}\right)+2 \left(y^2-\frac{9 y}{2}+\frac{81}{16}\right)=\fbox{$\frac{11}{2}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{x}{2}+\frac{1}{16}=\left(x-\frac{1}{4}\right)^2: \\
6 \fbox{$\left(x-\frac{1}{4}\right)^2$}+2 \left(y^2-\frac{9 y}{2}+\frac{81}{16}\right)=\frac{11}{2} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{9 y}{2}+\frac{81}{16}=\left(y-\frac{9}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 6 \left(x-\frac{1}{4}\right)^2+2 \fbox{$\left(y-\frac{9}{4}\right)^2$}=\frac{11}{2} \\
\end{array}
| khanacademy | amps |
Given the equation $x^2+7 x-4 y^2+6 y+5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-4 y^2+6 y+x^2+7 x+5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }5 \text{from }\text{both }\text{sides}: \\
-4 y^2+6 y+x^2+7 x=-5 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(x^2+7 x+\underline{\text{ }}\right)+\left(-4 y^2+6 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-4 y^2+6 y+\underline{\text{ }}\right)=-4 \left(y^2-\frac{3 y}{2}+\underline{\text{ }}\right): \\
\left(x^2+7 x+\underline{\text{ }}\right)+\fbox{$-4 \left(y^2-\frac{3 y}{2}+\underline{\text{ }}\right)$}=\underline{\text{ }}-5 \\
\end{array}
Step 5:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it, }\text{then }\text{add }\text{it }\text{to }\text{both }\text{sides.} \\
\text{Add }\left(\frac{7}{2}\right)^2=\frac{49}{4} \text{to }\text{both }\text{sides}: \\
\end{array}
Step 6:
\begin{array}{l}
\frac{49}{4}-5=\frac{29}{4}: \\
\left(x^2+7 x+\frac{49}{4}\right)-4 \left(y^2-\frac{3 y}{2}+\underline{\text{ }}\right)=\fbox{$\frac{29}{4}$} \\
\end{array}
Step 7:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-3}{2}}{2}\right)^2=\frac{9}{16} \text{on }\text{the }\text{left }\text{and }-4\times \frac{9}{16}=-\frac{9}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 8:
\begin{array}{l}
\frac{29}{4}-\frac{9}{4}=5: \\
\left(x^2+7 x+\frac{49}{4}\right)-4 \left(y^2-\frac{3 y}{2}+\frac{9}{16}\right)=\fbox{$5$} \\
\end{array}
Step 9:
\begin{array}{l}
x^2+7 x+\frac{49}{4}=\left(x+\frac{7}{2}\right)^2: \\
\fbox{$\left(x+\frac{7}{2}\right)^2$}-4 \left(y^2-\frac{3 y}{2}+\frac{9}{16}\right)=5 \\
\end{array}
Step 10:
\begin{array}{l}
y^2-\frac{3 y}{2}+\frac{9}{16}=\left(y-\frac{3}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & \left(x+\frac{7}{2}\right)^2-4 \fbox{$\left(y-\frac{3}{4}\right)^2$}=5 \\
\end{array}
| khanacademy | amps |
Given the equation $-3 x^2+6 x+y^2-9 y-8=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
y^2-9 y-3 x^2+6 x-8=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }8 \text{to }\text{both }\text{sides}: \\
y^2-9 y-3 x^2+6 x=8 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-3 x^2+6 x+\underline{\text{ }}\right)+\left(y^2-9 y+\underline{\text{ }}\right)=\underline{\text{ }}+8 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-3 x^2+6 x+\underline{\text{ }}\right)=-3 \left(x^2-2 x+\underline{\text{ }}\right): \\
\fbox{$-3 \left(x^2-2 x+\underline{\text{ }}\right)$}+\left(y^2-9 y+\underline{\text{ }}\right)=\underline{\text{ }}+8 \\
\end{array}
Step 5:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-2}{2}\right)^2=1 \text{on }\text{the }\text{left }\text{and }-3\times 1=-3 \text{on }\text{the }\text{right}: \\
\end{array}
Step 6:
\begin{array}{l}
8-3=5: \\
-3 \left(x^2-2 x+1\right)+\left(y^2-9 y+\underline{\text{ }}\right)=\fbox{$5$} \\
\end{array}
Step 7:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it, }\text{then }\text{add }\text{it }\text{to }\text{both }\text{sides.} \\
\text{Add }\left(\frac{-9}{2}\right)^2=\frac{81}{4} \text{to }\text{both }\text{sides}: \\
\end{array}
Step 8:
\begin{array}{l}
5+\frac{81}{4}=\frac{101}{4}: \\
-3 \left(x^2-2 x+1\right)+\left(y^2-9 y+\frac{81}{4}\right)=\fbox{$\frac{101}{4}$} \\
\end{array}
Step 9:
\begin{array}{l}
x^2-2 x+1=(x-1)^2: \\
-3 \fbox{$(x-1)^2$}+\left(y^2-9 y+\frac{81}{4}\right)=\frac{101}{4} \\
\end{array}
Step 10:
\begin{array}{l}
y^2-9 y+\frac{81}{4}=\left(y-\frac{9}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -3 (x-1)^2+\fbox{$\left(y-\frac{9}{2}\right)^2$}=\frac{101}{4} \\
\end{array}
| khanacademy | amps |
Given the equation $8 x^2-6 x+3 y^2+4=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
8 x^2-6 x+\left(3 y^2+4\right)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }4 \text{from }\text{both }\text{sides}: \\
3 y^2+8 x^2-6 x=-4 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{together }\text{on }\text{the }\text{left }\text{hand }\text{side, }\text{with }\text{a }\text{placeholder }\text{constant}: \\
\left(8 x^2-6 x+\underline{\text{ }}\right)+3 y^2=\underline{\text{ }}-4 \\
\end{array}
Step 4:
\begin{array}{l}
\left(8 x^2-6 x+\underline{\text{ }}\right)=8 \left(x^2-\frac{3 x}{4}+\underline{\text{ }}\right): \\
\fbox{$8 \left(x^2-\frac{3 x}{4}+\underline{\text{ }}\right)$}+3 y^2=\underline{\text{ }}-4 \\
\end{array}
Step 5:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-3}{4}}{2}\right)^2=\frac{9}{64} \text{on }\text{the }\text{left }\text{and }8\times \frac{9}{64}=\frac{9}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 6:
\begin{array}{l}
\frac{9}{8}-4=-\frac{23}{8}: \\
8 \left(x^2-\frac{3 x}{4}+\frac{9}{64}\right)+3 y^2=\fbox{$-\frac{23}{8}$} \\
\end{array}
Step 7:
\begin{array}{l}
x^2-\frac{3 x}{4}+\frac{9}{64}=\left(x-\frac{3}{8}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 8 \fbox{$\left(x-\frac{3}{8}\right)^2$}+3 y^2=-\frac{23}{8} \\
\end{array}
| khanacademy | amps |
Given the equation $4 x^2-x-3 y^2+2 y+6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-3 y^2+2 y+4 x^2-x+6=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }6 \text{from }\text{both }\text{sides}: \\
-3 y^2+2 y+4 x^2-x=-6 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(4 x^2-x+\underline{\text{ }}\right)+\left(-3 y^2+2 y+\underline{\text{ }}\right)=\underline{\text{ }}-6 \\
\end{array}
Step 4:
\begin{array}{l}
\left(4 x^2-x+\underline{\text{ }}\right)=4 \left(x^2-\frac{x}{4}+\underline{\text{ }}\right): \\
\fbox{$4 \left(x^2-\frac{x}{4}+\underline{\text{ }}\right)$}+\left(-3 y^2+2 y+\underline{\text{ }}\right)=\underline{\text{ }}-6 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-3 y^2+2 y+\underline{\text{ }}\right)=-3 \left(y^2-\frac{2 y}{3}+\underline{\text{ }}\right): \\
4 \left(x^2-\frac{x}{4}+\underline{\text{ }}\right)+\fbox{$-3 \left(y^2-\frac{2 y}{3}+\underline{\text{ }}\right)$}=\underline{\text{ }}-6 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-1}{4}}{2}\right)^2=\frac{1}{64} \text{on }\text{the }\text{left }\text{and }\frac{4}{64}=\frac{1}{16} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{1}{16}-6=-\frac{95}{16}: \\
4 \left(x^2-\frac{x}{4}+\frac{1}{64}\right)-3 \left(y^2-\frac{2 y}{3}+\underline{\text{ }}\right)=\fbox{$-\frac{95}{16}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-2}{3}}{2}\right)^2=\frac{1}{9} \text{on }\text{the }\text{left }\text{and }\frac{-3}{9}=-\frac{1}{3} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{95}{16}-\frac{1}{3}=-\frac{301}{48}: \\
4 \left(x^2-\frac{x}{4}+\frac{1}{64}\right)-3 \left(y^2-\frac{2 y}{3}+\frac{1}{9}\right)=\fbox{$-\frac{301}{48}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{x}{4}+\frac{1}{64}=\left(x-\frac{1}{8}\right)^2: \\
4 \fbox{$\left(x-\frac{1}{8}\right)^2$}-3 \left(y^2-\frac{2 y}{3}+\frac{1}{9}\right)=-\frac{301}{48} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{2 y}{3}+\frac{1}{9}=\left(y-\frac{1}{3}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 4 \left(x-\frac{1}{8}\right)^2-3 \fbox{$\left(y-\frac{1}{3}\right)^2$}=-\frac{301}{48} \\
\end{array}
| khanacademy | amps |
Given the equation $-8 x^2+5 x+4 y^2+2 y-6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
4 y^2+2 y-8 x^2+5 x-6=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }6 \text{to }\text{both }\text{sides}: \\
4 y^2+2 y-8 x^2+5 x=6 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-8 x^2+5 x+\underline{\text{ }}\right)+\left(4 y^2+2 y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-8 x^2+5 x+\underline{\text{ }}\right)=-8 \left(x^2-\frac{5 x}{8}+\underline{\text{ }}\right): \\
\fbox{$-8 \left(x^2-\frac{5 x}{8}+\underline{\text{ }}\right)$}+\left(4 y^2+2 y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\end{array}
Step 5:
\begin{array}{l}
\left(4 y^2+2 y+\underline{\text{ }}\right)=4 \left(y^2+\frac{y}{2}+\underline{\text{ }}\right): \\
-8 \left(x^2-\frac{5 x}{8}+\underline{\text{ }}\right)+\fbox{$4 \left(y^2+\frac{y}{2}+\underline{\text{ }}\right)$}=\underline{\text{ }}+6 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-5}{8}}{2}\right)^2=\frac{25}{256} \text{on }\text{the }\text{left }\text{and }-8\times \frac{25}{256}=-\frac{25}{32} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
6-\frac{25}{32}=\frac{167}{32}: \\
-8 \left(x^2-\frac{5 x}{8}+\frac{25}{256}\right)+4 \left(y^2+\frac{y}{2}+\underline{\text{ }}\right)=\fbox{$\frac{167}{32}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{2}}{2}\right)^2=\frac{1}{16} \text{on }\text{the }\text{left }\text{and }\frac{4}{16}=\frac{1}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{167}{32}+\frac{1}{4}=\frac{175}{32}: \\
-8 \left(x^2-\frac{5 x}{8}+\frac{25}{256}\right)+4 \left(y^2+\frac{y}{2}+\frac{1}{16}\right)=\fbox{$\frac{175}{32}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{5 x}{8}+\frac{25}{256}=\left(x-\frac{5}{16}\right)^2: \\
-8 \fbox{$\left(x-\frac{5}{16}\right)^2$}+4 \left(y^2+\frac{y}{2}+\frac{1}{16}\right)=\frac{175}{32} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{y}{2}+\frac{1}{16}=\left(y+\frac{1}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -8 \left(x-\frac{5}{16}\right)^2+4 \fbox{$\left(y+\frac{1}{4}\right)^2$}=\frac{175}{32} \\
\end{array}
| khanacademy | amps |
Given the equation $2 x^2+5 x+10 y^2+5 y+4=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
10 y^2+5 y+2 x^2+5 x+4=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }4 \text{from }\text{both }\text{sides}: \\
10 y^2+5 y+2 x^2+5 x=-4 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(2 x^2+5 x+\underline{\text{ }}\right)+\left(10 y^2+5 y+\underline{\text{ }}\right)=\underline{\text{ }}-4 \\
\end{array}
Step 4:
\begin{array}{l}
\left(2 x^2+5 x+\underline{\text{ }}\right)=2 \left(x^2+\frac{5 x}{2}+\underline{\text{ }}\right): \\
\fbox{$2 \left(x^2+\frac{5 x}{2}+\underline{\text{ }}\right)$}+\left(10 y^2+5 y+\underline{\text{ }}\right)=\underline{\text{ }}-4 \\
\end{array}
Step 5:
\begin{array}{l}
\left(10 y^2+5 y+\underline{\text{ }}\right)=10 \left(y^2+\frac{y}{2}+\underline{\text{ }}\right): \\
2 \left(x^2+\frac{5 x}{2}+\underline{\text{ }}\right)+\fbox{$10 \left(y^2+\frac{y}{2}+\underline{\text{ }}\right)$}=\underline{\text{ }}-4 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{5}{2}}{2}\right)^2=\frac{25}{16} \text{on }\text{the }\text{left }\text{and }2\times \frac{25}{16}=\frac{25}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{25}{8}-4=-\frac{7}{8}: \\
2 \left(x^2+\frac{5 x}{2}+\frac{25}{16}\right)+10 \left(y^2+\frac{y}{2}+\underline{\text{ }}\right)=\fbox{$-\frac{7}{8}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }\text{10 }\text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{2}}{2}\right)^2=\frac{1}{16} \text{on }\text{the }\text{left }\text{and }\frac{10}{16}=\frac{5}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{5}{8}-\frac{7}{8}=-\frac{1}{4}: \\
2 \left(x^2+\frac{5 x}{2}+\frac{25}{16}\right)+10 \left(y^2+\frac{y}{2}+\frac{1}{16}\right)=\fbox{$-\frac{1}{4}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{5 x}{2}+\frac{25}{16}=\left(x+\frac{5}{4}\right)^2: \\
2 \fbox{$\left(x+\frac{5}{4}\right)^2$}+10 \left(y^2+\frac{y}{2}+\frac{1}{16}\right)=-\frac{1}{4} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{y}{2}+\frac{1}{16}=\left(y+\frac{1}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 2 \left(x+\frac{5}{4}\right)^2+\text{10 }\fbox{$\left(y+\frac{1}{4}\right)^2$}=-\frac{1}{4} \\
\end{array}
| khanacademy | amps |
Given the equation $3 x^2-x+4 y^2-2 y-6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
4 y^2-2 y+3 x^2-x-6=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }6 \text{to }\text{both }\text{sides}: \\
4 y^2-2 y+3 x^2-x=6 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(3 x^2-x+\underline{\text{ }}\right)+\left(4 y^2-2 y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\end{array}
Step 4:
\begin{array}{l}
\left(3 x^2-x+\underline{\text{ }}\right)=3 \left(x^2-\frac{x}{3}+\underline{\text{ }}\right): \\
\fbox{$3 \left(x^2-\frac{x}{3}+\underline{\text{ }}\right)$}+\left(4 y^2-2 y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\end{array}
Step 5:
\begin{array}{l}
\left(4 y^2-2 y+\underline{\text{ }}\right)=4 \left(y^2-\frac{y}{2}+\underline{\text{ }}\right): \\
3 \left(x^2-\frac{x}{3}+\underline{\text{ }}\right)+\fbox{$4 \left(y^2-\frac{y}{2}+\underline{\text{ }}\right)$}=\underline{\text{ }}+6 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-1}{3}}{2}\right)^2=\frac{1}{36} \text{on }\text{the }\text{left }\text{and }\frac{3}{36}=\frac{1}{12} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
6+\frac{1}{12}=\frac{73}{12}: \\
3 \left(x^2-\frac{x}{3}+\frac{1}{36}\right)+4 \left(y^2-\frac{y}{2}+\underline{\text{ }}\right)=\fbox{$\frac{73}{12}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-1}{2}}{2}\right)^2=\frac{1}{16} \text{on }\text{the }\text{left }\text{and }\frac{4}{16}=\frac{1}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{73}{12}+\frac{1}{4}=\frac{19}{3}: \\
3 \left(x^2-\frac{x}{3}+\frac{1}{36}\right)+4 \left(y^2-\frac{y}{2}+\frac{1}{16}\right)=\fbox{$\frac{19}{3}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{x}{3}+\frac{1}{36}=\left(x-\frac{1}{6}\right)^2: \\
3 \fbox{$\left(x-\frac{1}{6}\right)^2$}+4 \left(y^2-\frac{y}{2}+\frac{1}{16}\right)=\frac{19}{3} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{y}{2}+\frac{1}{16}=\left(y-\frac{1}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 3 \left(x-\frac{1}{6}\right)^2+4 \fbox{$\left(y-\frac{1}{4}\right)^2$}=\frac{19}{3} \\
\end{array}
| khanacademy | amps |
Given the equation $10 x^2+8 x-3 y^2+6 y+2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-3 y^2+6 y+10 x^2+8 x+2=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }2 \text{from }\text{both }\text{sides}: \\
-3 y^2+6 y+10 x^2+8 x=-2 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(10 x^2+8 x+\underline{\text{ }}\right)+\left(-3 y^2+6 y+\underline{\text{ }}\right)=\underline{\text{ }}-2 \\
\end{array}
Step 4:
\begin{array}{l}
\left(10 x^2+8 x+\underline{\text{ }}\right)=10 \left(x^2+\frac{4 x}{5}+\underline{\text{ }}\right): \\
\fbox{$10 \left(x^2+\frac{4 x}{5}+\underline{\text{ }}\right)$}+\left(-3 y^2+6 y+\underline{\text{ }}\right)=\underline{\text{ }}-2 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-3 y^2+6 y+\underline{\text{ }}\right)=-3 \left(y^2-2 y+\underline{\text{ }}\right): \\
10 \left(x^2+\frac{4 x}{5}+\underline{\text{ }}\right)+\fbox{$-3 \left(y^2-2 y+\underline{\text{ }}\right)$}=\underline{\text{ }}-2 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }\text{10 }\text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{4}{5}}{2}\right)^2=\frac{4}{25} \text{on }\text{the }\text{left }\text{and }10\times \frac{4}{25}=\frac{8}{5} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{8}{5}-2=-\frac{2}{5}: \\
10 \left(x^2+\frac{4 x}{5}+\frac{4}{25}\right)-3 \left(y^2-2 y+\underline{\text{ }}\right)=\fbox{$-\frac{2}{5}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-2}{2}\right)^2=1 \text{on }\text{the }\text{left }\text{and }-3\times 1=-3 \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{2}{5}-3=-\frac{17}{5}: \\
10 \left(x^2+\frac{4 x}{5}+\frac{4}{25}\right)-3 \left(y^2-2 y+1\right)=\fbox{$-\frac{17}{5}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{4 x}{5}+\frac{4}{25}=\left(x+\frac{2}{5}\right)^2: \\
\text{10 }\fbox{$\left(x+\frac{2}{5}\right)^2$}-3 \left(y^2-2 y+1\right)=-\frac{17}{5} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-2 y+1=(y-1)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 10 \left(x+\frac{2}{5}\right)^2-3 \fbox{$(y-1)^2$}=-\frac{17}{5} \\
\end{array}
| khanacademy | amps |
Given the equation $-6 x^2+x+7 y^2-6 y+3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
7 y^2-6 y-6 x^2+x+3=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }3 \text{from }\text{both }\text{sides}: \\
7 y^2-6 y-6 x^2+x=-3 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-6 x^2+x+\underline{\text{ }}\right)+\left(7 y^2-6 y+\underline{\text{ }}\right)=\underline{\text{ }}-3 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-6 x^2+x+\underline{\text{ }}\right)=-6 \left(x^2-\frac{x}{6}+\underline{\text{ }}\right): \\
\fbox{$-6 \left(x^2-\frac{x}{6}+\underline{\text{ }}\right)$}+\left(7 y^2-6 y+\underline{\text{ }}\right)=\underline{\text{ }}-3 \\
\end{array}
Step 5:
\begin{array}{l}
\left(7 y^2-6 y+\underline{\text{ }}\right)=7 \left(y^2-\frac{6 y}{7}+\underline{\text{ }}\right): \\
-6 \left(x^2-\frac{x}{6}+\underline{\text{ }}\right)+\fbox{$7 \left(y^2-\frac{6 y}{7}+\underline{\text{ }}\right)$}=\underline{\text{ }}-3 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-1}{6}}{2}\right)^2=\frac{1}{144} \text{on }\text{the }\text{left }\text{and }\frac{-6}{144}=-\frac{1}{24} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-3-\frac{1}{24}=-\frac{73}{24}: \\
-6 \left(x^2-\frac{x}{6}+\frac{1}{144}\right)+7 \left(y^2-\frac{6 y}{7}+\underline{\text{ }}\right)=\fbox{$-\frac{73}{24}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-6}{7}}{2}\right)^2=\frac{9}{49} \text{on }\text{the }\text{left }\text{and }7\times \frac{9}{49}=\frac{9}{7} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{9}{7}-\frac{73}{24}=-\frac{295}{168}: \\
-6 \left(x^2-\frac{x}{6}+\frac{1}{144}\right)+7 \left(y^2-\frac{6 y}{7}+\frac{9}{49}\right)=\fbox{$-\frac{295}{168}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{x}{6}+\frac{1}{144}=\left(x-\frac{1}{12}\right)^2: \\
-6 \fbox{$\left(x-\frac{1}{12}\right)^2$}+7 \left(y^2-\frac{6 y}{7}+\frac{9}{49}\right)=-\frac{295}{168} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{6 y}{7}+\frac{9}{49}=\left(y-\frac{3}{7}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -6 \left(x-\frac{1}{12}\right)^2+7 \fbox{$\left(y-\frac{3}{7}\right)^2$}=-\frac{295}{168} \\
\end{array}
| khanacademy | amps |
Given the equation $-9 x^2+2 x+8 y^2-10 y+7=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
8 y^2-10 y-9 x^2+2 x+7=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }7 \text{from }\text{both }\text{sides}: \\
8 y^2-10 y-9 x^2+2 x=-7 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-9 x^2+2 x+\underline{\text{ }}\right)+\left(8 y^2-10 y+\underline{\text{ }}\right)=\underline{\text{ }}-7 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-9 x^2+2 x+\underline{\text{ }}\right)=-9 \left(x^2-\frac{2 x}{9}+\underline{\text{ }}\right): \\
\fbox{$-9 \left(x^2-\frac{2 x}{9}+\underline{\text{ }}\right)$}+\left(8 y^2-10 y+\underline{\text{ }}\right)=\underline{\text{ }}-7 \\
\end{array}
Step 5:
\begin{array}{l}
\left(8 y^2-10 y+\underline{\text{ }}\right)=8 \left(y^2-\frac{5 y}{4}+\underline{\text{ }}\right): \\
-9 \left(x^2-\frac{2 x}{9}+\underline{\text{ }}\right)+\fbox{$8 \left(y^2-\frac{5 y}{4}+\underline{\text{ }}\right)$}=\underline{\text{ }}-7 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-2}{9}}{2}\right)^2=\frac{1}{81} \text{on }\text{the }\text{left }\text{and }\frac{-9}{81}=-\frac{1}{9} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-7-\frac{1}{9}=-\frac{64}{9}: \\
-9 \left(x^2-\frac{2 x}{9}+\frac{1}{81}\right)+8 \left(y^2-\frac{5 y}{4}+\underline{\text{ }}\right)=\fbox{$-\frac{64}{9}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-5}{4}}{2}\right)^2=\frac{25}{64} \text{on }\text{the }\text{left }\text{and }8\times \frac{25}{64}=\frac{25}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{25}{8}-\frac{64}{9}=-\frac{287}{72}: \\
-9 \left(x^2-\frac{2 x}{9}+\frac{1}{81}\right)+8 \left(y^2-\frac{5 y}{4}+\frac{25}{64}\right)=\fbox{$-\frac{287}{72}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{2 x}{9}+\frac{1}{81}=\left(x-\frac{1}{9}\right)^2: \\
-9 \fbox{$\left(x-\frac{1}{9}\right)^2$}+8 \left(y^2-\frac{5 y}{4}+\frac{25}{64}\right)=-\frac{287}{72} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{5 y}{4}+\frac{25}{64}=\left(y-\frac{5}{8}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -9 \left(x-\frac{1}{9}\right)^2+8 \fbox{$\left(y-\frac{5}{8}\right)^2$}=-\frac{287}{72} \\
\end{array}
| khanacademy | amps |
Given the equation $5 x^2+8 x+3 y^2+6 y-10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
3 y^2+6 y+5 x^2+8 x-10=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }\text{10 }\text{to }\text{both }\text{sides}: \\
3 y^2+6 y+5 x^2+8 x=10 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(5 x^2+8 x+\underline{\text{ }}\right)+\left(3 y^2+6 y+\underline{\text{ }}\right)=\underline{\text{ }}+10 \\
\end{array}
Step 4:
\begin{array}{l}
\left(5 x^2+8 x+\underline{\text{ }}\right)=5 \left(x^2+\frac{8 x}{5}+\underline{\text{ }}\right): \\
\fbox{$5 \left(x^2+\frac{8 x}{5}+\underline{\text{ }}\right)$}+\left(3 y^2+6 y+\underline{\text{ }}\right)=\underline{\text{ }}+10 \\
\end{array}
Step 5:
\begin{array}{l}
\left(3 y^2+6 y+\underline{\text{ }}\right)=3 \left(y^2+2 y+\underline{\text{ }}\right): \\
5 \left(x^2+\frac{8 x}{5}+\underline{\text{ }}\right)+\fbox{$3 \left(y^2+2 y+\underline{\text{ }}\right)$}=\underline{\text{ }}+10 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{8}{5}}{2}\right)^2=\frac{16}{25} \text{on }\text{the }\text{left }\text{and }5\times \frac{16}{25}=\frac{16}{5} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
10+\frac{16}{5}=\frac{66}{5}: \\
5 \left(x^2+\frac{8 x}{5}+\frac{16}{25}\right)+3 \left(y^2+2 y+\underline{\text{ }}\right)=\fbox{$\frac{66}{5}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{2}{2}\right)^2=1 \text{on }\text{the }\text{left }\text{and }3\times 1=3 \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{66}{5}+3=\frac{81}{5}: \\
5 \left(x^2+\frac{8 x}{5}+\frac{16}{25}\right)+3 \left(y^2+2 y+1\right)=\fbox{$\frac{81}{5}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{8 x}{5}+\frac{16}{25}=\left(x+\frac{4}{5}\right)^2: \\
5 \fbox{$\left(x+\frac{4}{5}\right)^2$}+3 \left(y^2+2 y+1\right)=\frac{81}{5} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+2 y+1=(y+1)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 5 \left(x+\frac{4}{5}\right)^2+3 \fbox{$(y+1)^2$}=\frac{81}{5} \\
\end{array}
| khanacademy | amps |
Given the equation $-8 x^2-2 x-9 y^2-2 y+8=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-9 y^2-2 y-8 x^2-2 x+8=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }8 \text{from }\text{both }\text{sides}: \\
-9 y^2-2 y-8 x^2-2 x=-8 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-8 x^2-2 x+\underline{\text{ }}\right)+\left(-9 y^2-2 y+\underline{\text{ }}\right)=\underline{\text{ }}-8 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-8 x^2-2 x+\underline{\text{ }}\right)=-8 \left(x^2+\frac{x}{4}+\underline{\text{ }}\right): \\
\fbox{$-8 \left(x^2+\frac{x}{4}+\underline{\text{ }}\right)$}+\left(-9 y^2-2 y+\underline{\text{ }}\right)=\underline{\text{ }}-8 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-9 y^2-2 y+\underline{\text{ }}\right)=-9 \left(y^2+\frac{2 y}{9}+\underline{\text{ }}\right): \\
-8 \left(x^2+\frac{x}{4}+\underline{\text{ }}\right)+\fbox{$-9 \left(y^2+\frac{2 y}{9}+\underline{\text{ }}\right)$}=\underline{\text{ }}-8 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{4}}{2}\right)^2=\frac{1}{64} \text{on }\text{the }\text{left }\text{and }\frac{-8}{64}=-\frac{1}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-8-\frac{1}{8}=-\frac{65}{8}: \\
-8 \left(x^2+\frac{x}{4}+\frac{1}{64}\right)-9 \left(y^2+\frac{2 y}{9}+\underline{\text{ }}\right)=\fbox{$-\frac{65}{8}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{2}{9}}{2}\right)^2=\frac{1}{81} \text{on }\text{the }\text{left }\text{and }\frac{-9}{81}=-\frac{1}{9} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{65}{8}-\frac{1}{9}=-\frac{593}{72}: \\
-8 \left(x^2+\frac{x}{4}+\frac{1}{64}\right)-9 \left(y^2+\frac{2 y}{9}+\frac{1}{81}\right)=\fbox{$-\frac{593}{72}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{x}{4}+\frac{1}{64}=\left(x+\frac{1}{8}\right)^2: \\
-8 \fbox{$\left(x+\frac{1}{8}\right)^2$}-9 \left(y^2+\frac{2 y}{9}+\frac{1}{81}\right)=-\frac{593}{72} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{2 y}{9}+\frac{1}{81}=\left(y+\frac{1}{9}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -8 \left(x+\frac{1}{8}\right)^2-9 \fbox{$\left(y+\frac{1}{9}\right)^2$}=-\frac{593}{72} \\
\end{array}
| khanacademy | amps |
Given the equation $-9 x^2+10 x-4 y^2-3 y-8=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-4 y^2-3 y-9 x^2+10 x-8=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }8 \text{to }\text{both }\text{sides}: \\
-4 y^2-3 y-9 x^2+10 x=8 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-9 x^2+10 x+\underline{\text{ }}\right)+\left(-4 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}+8 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-9 x^2+10 x+\underline{\text{ }}\right)=-9 \left(x^2-\frac{10 x}{9}+\underline{\text{ }}\right): \\
\fbox{$-9 \left(x^2-\frac{10 x}{9}+\underline{\text{ }}\right)$}+\left(-4 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}+8 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-4 y^2-3 y+\underline{\text{ }}\right)=-4 \left(y^2+\frac{3 y}{4}+\underline{\text{ }}\right): \\
-9 \left(x^2-\frac{10 x}{9}+\underline{\text{ }}\right)+\fbox{$-4 \left(y^2+\frac{3 y}{4}+\underline{\text{ }}\right)$}=\underline{\text{ }}+8 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-10}{9}}{2}\right)^2=\frac{25}{81} \text{on }\text{the }\text{left }\text{and }-9\times \frac{25}{81}=-\frac{25}{9} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
8-\frac{25}{9}=\frac{47}{9}: \\
-9 \left(x^2-\frac{10 x}{9}+\frac{25}{81}\right)-4 \left(y^2+\frac{3 y}{4}+\underline{\text{ }}\right)=\fbox{$\frac{47}{9}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{3}{4}}{2}\right)^2=\frac{9}{64} \text{on }\text{the }\text{left }\text{and }-4\times \frac{9}{64}=-\frac{9}{16} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{47}{9}-\frac{9}{16}=\frac{671}{144}: \\
-9 \left(x^2-\frac{10 x}{9}+\frac{25}{81}\right)-4 \left(y^2+\frac{3 y}{4}+\frac{9}{64}\right)=\fbox{$\frac{671}{144}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{10 x}{9}+\frac{25}{81}=\left(x-\frac{5}{9}\right)^2: \\
-9 \fbox{$\left(x-\frac{5}{9}\right)^2$}-4 \left(y^2+\frac{3 y}{4}+\frac{9}{64}\right)=\frac{671}{144} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{3 y}{4}+\frac{9}{64}=\left(y+\frac{3}{8}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -9 \left(x-\frac{5}{9}\right)^2-4 \fbox{$\left(y+\frac{3}{8}\right)^2$}=\frac{671}{144} \\
\end{array}
| khanacademy | amps |
Given the equation $-x^2-7 x-9 y^2+10 y+6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-9 y^2+10 y-x^2-7 x+6=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }6 \text{from }\text{both }\text{sides}: \\
-9 y^2+10 y-x^2-7 x=-6 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-x^2-7 x+\underline{\text{ }}\right)+\left(-9 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}-6 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-x^2-7 x+\underline{\text{ }}\right)=-\left(x^2+7 x+\underline{\text{ }}\right): \\
\fbox{$-\left(x^2+7 x+\underline{\text{ }}\right)$}+\left(-9 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}-6 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-9 y^2+10 y+\underline{\text{ }}\right)=-9 \left(y^2-\frac{10 y}{9}+\underline{\text{ }}\right): \\
-\left(x^2+7 x+\underline{\text{ }}\right)+\fbox{$-9 \left(y^2-\frac{10 y}{9}+\underline{\text{ }}\right)$}=\underline{\text{ }}-6 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-1 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{7}{2}\right)^2=\frac{49}{4} \text{on }\text{the }\text{left }\text{and }-\frac{49}{4}=-\frac{49}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-6-\frac{49}{4}=-\frac{73}{4}: \\
-\left(x^2+7 x+\frac{49}{4}\right)-9 \left(y^2-\frac{10 y}{9}+\underline{\text{ }}\right)=\fbox{$-\frac{73}{4}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-10}{9}}{2}\right)^2=\frac{25}{81} \text{on }\text{the }\text{left }\text{and }-9\times \frac{25}{81}=-\frac{25}{9} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{73}{4}-\frac{25}{9}=-\frac{757}{36}: \\
-\left(x^2+7 x+\frac{49}{4}\right)-9 \left(y^2-\frac{10 y}{9}+\frac{25}{81}\right)=\fbox{$-\frac{757}{36}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+7 x+\frac{49}{4}=\left(x+\frac{7}{2}\right)^2: \\
-\fbox{$\left(x+\frac{7}{2}\right)^2$}-9 \left(y^2-\frac{10 y}{9}+\frac{25}{81}\right)=-\frac{757}{36} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{10 y}{9}+\frac{25}{81}=\left(y-\frac{5}{9}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -\left(x+\frac{7}{2}\right)^2-9 \fbox{$\left(y-\frac{5}{9}\right)^2$}=-\frac{757}{36} \\
\end{array}
| khanacademy | amps |
Given the equation $3 x^2-4 x-y^2+2 y+1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-y^2+2 y+3 x^2-4 x+1=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }1 \text{from }\text{both }\text{sides}: \\
-y^2+2 y+3 x^2-4 x=-1 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(3 x^2-4 x+\underline{\text{ }}\right)+\left(-y^2+2 y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\end{array}
Step 4:
\begin{array}{l}
\left(3 x^2-4 x+\underline{\text{ }}\right)=3 \left(x^2-\frac{4 x}{3}+\underline{\text{ }}\right): \\
\fbox{$3 \left(x^2-\frac{4 x}{3}+\underline{\text{ }}\right)$}+\left(-y^2+2 y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-y^2+2 y+\underline{\text{ }}\right)=-\left(y^2-2 y+\underline{\text{ }}\right): \\
3 \left(x^2-\frac{4 x}{3}+\underline{\text{ }}\right)+\fbox{$-\left(y^2-2 y+\underline{\text{ }}\right)$}=\underline{\text{ }}-1 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-4}{3}}{2}\right)^2=\frac{4}{9} \text{on }\text{the }\text{left }\text{and }3\times \frac{4}{9}=\frac{4}{3} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{4}{3}-1=\frac{1}{3}: \\
3 \left(x^2-\frac{4 x}{3}+\frac{4}{9}\right)-\left(y^2-2 y+\underline{\text{ }}\right)=\fbox{$\frac{1}{3}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-1 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-2}{2}\right)^2=1 \text{on }\text{the }\text{left }\text{and }-1=-1 \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{1}{3}-1=-\frac{2}{3}: \\
3 \left(x^2-\frac{4 x}{3}+\frac{4}{9}\right)-\left(y^2-2 y+1\right)=\fbox{$-\frac{2}{3}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{4 x}{3}+\frac{4}{9}=\left(x-\frac{2}{3}\right)^2: \\
3 \fbox{$\left(x-\frac{2}{3}\right)^2$}-\left(y^2-2 y+1\right)=-\frac{2}{3} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-2 y+1=(y-1)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 3 \left(x-\frac{2}{3}\right)^2-\fbox{$(y-1)^2$}=-\frac{2}{3} \\
\end{array}
| khanacademy | amps |
Given the equation $10 x^2+8 x-6 y^2-5 y-7=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-6 y^2-5 y+10 x^2+8 x-7=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }7 \text{to }\text{both }\text{sides}: \\
-6 y^2-5 y+10 x^2+8 x=7 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(10 x^2+8 x+\underline{\text{ }}\right)+\left(-6 y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}+7 \\
\end{array}
Step 4:
\begin{array}{l}
\left(10 x^2+8 x+\underline{\text{ }}\right)=10 \left(x^2+\frac{4 x}{5}+\underline{\text{ }}\right): \\
\fbox{$10 \left(x^2+\frac{4 x}{5}+\underline{\text{ }}\right)$}+\left(-6 y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}+7 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-6 y^2-5 y+\underline{\text{ }}\right)=-6 \left(y^2+\frac{5 y}{6}+\underline{\text{ }}\right): \\
10 \left(x^2+\frac{4 x}{5}+\underline{\text{ }}\right)+\fbox{$-6 \left(y^2+\frac{5 y}{6}+\underline{\text{ }}\right)$}=\underline{\text{ }}+7 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }\text{10 }\text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{4}{5}}{2}\right)^2=\frac{4}{25} \text{on }\text{the }\text{left }\text{and }10\times \frac{4}{25}=\frac{8}{5} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
7+\frac{8}{5}=\frac{43}{5}: \\
10 \left(x^2+\frac{4 x}{5}+\frac{4}{25}\right)-6 \left(y^2+\frac{5 y}{6}+\underline{\text{ }}\right)=\fbox{$\frac{43}{5}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{5}{6}}{2}\right)^2=\frac{25}{144} \text{on }\text{the }\text{left }\text{and }-6\times \frac{25}{144}=-\frac{25}{24} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{43}{5}-\frac{25}{24}=\frac{907}{120}: \\
10 \left(x^2+\frac{4 x}{5}+\frac{4}{25}\right)-6 \left(y^2+\frac{5 y}{6}+\frac{25}{144}\right)=\fbox{$\frac{907}{120}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{4 x}{5}+\frac{4}{25}=\left(x+\frac{2}{5}\right)^2: \\
\text{10 }\fbox{$\left(x+\frac{2}{5}\right)^2$}-6 \left(y^2+\frac{5 y}{6}+\frac{25}{144}\right)=\frac{907}{120} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{5 y}{6}+\frac{25}{144}=\left(y+\frac{5}{12}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 10 \left(x+\frac{2}{5}\right)^2-6 \fbox{$\left(y+\frac{5}{12}\right)^2$}=\frac{907}{120} \\
\end{array}
| khanacademy | amps |
Given the equation $-7 x^2-4 x+7 y^2-9 y+6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
7 y^2-9 y-7 x^2-4 x+6=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }6 \text{from }\text{both }\text{sides}: \\
7 y^2-9 y-7 x^2-4 x=-6 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-7 x^2-4 x+\underline{\text{ }}\right)+\left(7 y^2-9 y+\underline{\text{ }}\right)=\underline{\text{ }}-6 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-7 x^2-4 x+\underline{\text{ }}\right)=-7 \left(x^2+\frac{4 x}{7}+\underline{\text{ }}\right): \\
\fbox{$-7 \left(x^2+\frac{4 x}{7}+\underline{\text{ }}\right)$}+\left(7 y^2-9 y+\underline{\text{ }}\right)=\underline{\text{ }}-6 \\
\end{array}
Step 5:
\begin{array}{l}
\left(7 y^2-9 y+\underline{\text{ }}\right)=7 \left(y^2-\frac{9 y}{7}+\underline{\text{ }}\right): \\
-7 \left(x^2+\frac{4 x}{7}+\underline{\text{ }}\right)+\fbox{$7 \left(y^2-\frac{9 y}{7}+\underline{\text{ }}\right)$}=\underline{\text{ }}-6 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{4}{7}}{2}\right)^2=\frac{4}{49} \text{on }\text{the }\text{left }\text{and }-7\times \frac{4}{49}=-\frac{4}{7} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-6-\frac{4}{7}=-\frac{46}{7}: \\
-7 \left(x^2+\frac{4 x}{7}+\frac{4}{49}\right)+7 \left(y^2-\frac{9 y}{7}+\underline{\text{ }}\right)=\fbox{$-\frac{46}{7}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-9}{7}}{2}\right)^2=\frac{81}{196} \text{on }\text{the }\text{left }\text{and }7\times \frac{81}{196}=\frac{81}{28} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{81}{28}-\frac{46}{7}=-\frac{103}{28}: \\
-7 \left(x^2+\frac{4 x}{7}+\frac{4}{49}\right)+7 \left(y^2-\frac{9 y}{7}+\frac{81}{196}\right)=\fbox{$-\frac{103}{28}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{4 x}{7}+\frac{4}{49}=\left(x+\frac{2}{7}\right)^2: \\
-7 \fbox{$\left(x+\frac{2}{7}\right)^2$}+7 \left(y^2-\frac{9 y}{7}+\frac{81}{196}\right)=-\frac{103}{28} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{9 y}{7}+\frac{81}{196}=\left(y-\frac{9}{14}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -7 \left(x+\frac{2}{7}\right)^2+7 \fbox{$\left(y-\frac{9}{14}\right)^2$}=-\frac{103}{28} \\
\end{array}
| khanacademy | amps |
Given the equation $8 x^2+2 x+6 y^2+6 y-3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
6 y^2+6 y+8 x^2+2 x-3=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }3 \text{to }\text{both }\text{sides}: \\
6 y^2+6 y+8 x^2+2 x=3 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(8 x^2+2 x+\underline{\text{ }}\right)+\left(6 y^2+6 y+\underline{\text{ }}\right)=\underline{\text{ }}+3 \\
\end{array}
Step 4:
\begin{array}{l}
\left(8 x^2+2 x+\underline{\text{ }}\right)=8 \left(x^2+\frac{x}{4}+\underline{\text{ }}\right): \\
\fbox{$8 \left(x^2+\frac{x}{4}+\underline{\text{ }}\right)$}+\left(6 y^2+6 y+\underline{\text{ }}\right)=\underline{\text{ }}+3 \\
\end{array}
Step 5:
\begin{array}{l}
\left(6 y^2+6 y+\underline{\text{ }}\right)=6 \left(y^2+y+\underline{\text{ }}\right): \\
8 \left(x^2+\frac{x}{4}+\underline{\text{ }}\right)+\fbox{$6 \left(y^2+y+\underline{\text{ }}\right)$}=\underline{\text{ }}+3 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{4}}{2}\right)^2=\frac{1}{64} \text{on }\text{the }\text{left }\text{and }\frac{8}{64}=\frac{1}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
3+\frac{1}{8}=\frac{25}{8}: \\
8 \left(x^2+\frac{x}{4}+\frac{1}{64}\right)+6 \left(y^2+y+\underline{\text{ }}\right)=\fbox{$\frac{25}{8}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{6}{4}=\frac{3}{2} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{25}{8}+\frac{3}{2}=\frac{37}{8}: \\
8 \left(x^2+\frac{x}{4}+\frac{1}{64}\right)+6 \left(y^2+y+\frac{1}{4}\right)=\fbox{$\frac{37}{8}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{x}{4}+\frac{1}{64}=\left(x+\frac{1}{8}\right)^2: \\
8 \fbox{$\left(x+\frac{1}{8}\right)^2$}+6 \left(y^2+y+\frac{1}{4}\right)=\frac{37}{8} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+y+\frac{1}{4}=\left(y+\frac{1}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 8 \left(x+\frac{1}{8}\right)^2+6 \fbox{$\left(y+\frac{1}{2}\right)^2$}=\frac{37}{8} \\
\end{array}
| khanacademy | amps |
Given the equation $-6 x^2+x-3 y^2+8 y+7=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-3 y^2+8 y-6 x^2+x+7=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }7 \text{from }\text{both }\text{sides}: \\
-3 y^2+8 y-6 x^2+x=-7 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-6 x^2+x+\underline{\text{ }}\right)+\left(-3 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}-7 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-6 x^2+x+\underline{\text{ }}\right)=-6 \left(x^2-\frac{x}{6}+\underline{\text{ }}\right): \\
\fbox{$-6 \left(x^2-\frac{x}{6}+\underline{\text{ }}\right)$}+\left(-3 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}-7 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-3 y^2+8 y+\underline{\text{ }}\right)=-3 \left(y^2-\frac{8 y}{3}+\underline{\text{ }}\right): \\
-6 \left(x^2-\frac{x}{6}+\underline{\text{ }}\right)+\fbox{$-3 \left(y^2-\frac{8 y}{3}+\underline{\text{ }}\right)$}=\underline{\text{ }}-7 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-1}{6}}{2}\right)^2=\frac{1}{144} \text{on }\text{the }\text{left }\text{and }\frac{-6}{144}=-\frac{1}{24} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-7-\frac{1}{24}=-\frac{169}{24}: \\
-6 \left(x^2-\frac{x}{6}+\frac{1}{144}\right)-3 \left(y^2-\frac{8 y}{3}+\underline{\text{ }}\right)=\fbox{$-\frac{169}{24}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-8}{3}}{2}\right)^2=\frac{16}{9} \text{on }\text{the }\text{left }\text{and }-3\times \frac{16}{9}=-\frac{16}{3} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{169}{24}-\frac{16}{3}=-\frac{99}{8}: \\
-6 \left(x^2-\frac{x}{6}+\frac{1}{144}\right)-3 \left(y^2-\frac{8 y}{3}+\frac{16}{9}\right)=\fbox{$-\frac{99}{8}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{x}{6}+\frac{1}{144}=\left(x-\frac{1}{12}\right)^2: \\
-6 \fbox{$\left(x-\frac{1}{12}\right)^2$}-3 \left(y^2-\frac{8 y}{3}+\frac{16}{9}\right)=-\frac{99}{8} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{8 y}{3}+\frac{16}{9}=\left(y-\frac{4}{3}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -6 \left(x-\frac{1}{12}\right)^2-3 \fbox{$\left(y-\frac{4}{3}\right)^2$}=-\frac{99}{8} \\
\end{array}
| khanacademy | amps |
Given the equation $4 x^2+3 x+y=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
y+4 x^2+3 x=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }y \text{from }\text{both }\text{sides}: \\
4 x^2+3 x=-y \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{together }\text{on }\text{the }\text{left }\text{hand }\text{side, }\text{with }\text{a }\text{placeholder }\text{constant}: \\
\left(4 x^2+3 x+\underline{\text{ }}\right)=\underline{\text{ }}-y \\
\end{array}
Step 4:
\begin{array}{l}
\left(4 x^2+3 x+\underline{\text{ }}\right)=4 \left(x^2+\frac{3 x}{4}+\underline{\text{ }}\right): \\
\fbox{$4 \left(x^2+\frac{3 x}{4}+\underline{\text{ }}\right)$}=\underline{\text{ }}-y \\
\end{array}
Step 5:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{3}{4}}{2}\right)^2=\frac{9}{64} \text{on }\text{the }\text{left }\text{and }4\times \frac{9}{64}=\frac{9}{16} \text{on }\text{the }\text{right}: \\
\end{array}
Step 6:
\begin{array}{l}
x^2+\frac{3 x}{4}+\frac{9}{64}=\left(x+\frac{3}{8}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 4 \fbox{$\left(x+\frac{3}{8}\right)^2$}=\frac{9}{16}-y \\
\end{array}
| khanacademy | amps |
Given the equation $-7 x^2-5 x+10 y^2+4 y+9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
10 y^2+4 y-7 x^2-5 x+9=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }9 \text{from }\text{both }\text{sides}: \\
10 y^2+4 y-7 x^2-5 x=-9 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-7 x^2-5 x+\underline{\text{ }}\right)+\left(10 y^2+4 y+\underline{\text{ }}\right)=\underline{\text{ }}-9 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-7 x^2-5 x+\underline{\text{ }}\right)=-7 \left(x^2+\frac{5 x}{7}+\underline{\text{ }}\right): \\
\fbox{$-7 \left(x^2+\frac{5 x}{7}+\underline{\text{ }}\right)$}+\left(10 y^2+4 y+\underline{\text{ }}\right)=\underline{\text{ }}-9 \\
\end{array}
Step 5:
\begin{array}{l}
\left(10 y^2+4 y+\underline{\text{ }}\right)=10 \left(y^2+\frac{2 y}{5}+\underline{\text{ }}\right): \\
-7 \left(x^2+\frac{5 x}{7}+\underline{\text{ }}\right)+\fbox{$10 \left(y^2+\frac{2 y}{5}+\underline{\text{ }}\right)$}=\underline{\text{ }}-9 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{5}{7}}{2}\right)^2=\frac{25}{196} \text{on }\text{the }\text{left }\text{and }-7\times \frac{25}{196}=-\frac{25}{28} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-9-\frac{25}{28}=-\frac{277}{28}: \\
-7 \left(x^2+\frac{5 x}{7}+\frac{25}{196}\right)+10 \left(y^2+\frac{2 y}{5}+\underline{\text{ }}\right)=\fbox{$-\frac{277}{28}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }\text{10 }\text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{2}{5}}{2}\right)^2=\frac{1}{25} \text{on }\text{the }\text{left }\text{and }\frac{10}{25}=\frac{2}{5} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{2}{5}-\frac{277}{28}=-\frac{1329}{140}: \\
-7 \left(x^2+\frac{5 x}{7}+\frac{25}{196}\right)+10 \left(y^2+\frac{2 y}{5}+\frac{1}{25}\right)=\fbox{$-\frac{1329}{140}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{5 x}{7}+\frac{25}{196}=\left(x+\frac{5}{14}\right)^2: \\
-7 \fbox{$\left(x+\frac{5}{14}\right)^2$}+10 \left(y^2+\frac{2 y}{5}+\frac{1}{25}\right)=-\frac{1329}{140} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{2 y}{5}+\frac{1}{25}=\left(y+\frac{1}{5}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -7 \left(x+\frac{5}{14}\right)^2+\text{10 }\fbox{$\left(y+\frac{1}{5}\right)^2$}=-\frac{1329}{140} \\
\end{array}
| khanacademy | amps |
Given the equation $3 x^2-4 x+2 y^2+3 y+4=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
2 y^2+3 y+3 x^2-4 x+4=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }4 \text{from }\text{both }\text{sides}: \\
2 y^2+3 y+3 x^2-4 x=-4 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(3 x^2-4 x+\underline{\text{ }}\right)+\left(2 y^2+3 y+\underline{\text{ }}\right)=\underline{\text{ }}-4 \\
\end{array}
Step 4:
\begin{array}{l}
\left(3 x^2-4 x+\underline{\text{ }}\right)=3 \left(x^2-\frac{4 x}{3}+\underline{\text{ }}\right): \\
\fbox{$3 \left(x^2-\frac{4 x}{3}+\underline{\text{ }}\right)$}+\left(2 y^2+3 y+\underline{\text{ }}\right)=\underline{\text{ }}-4 \\
\end{array}
Step 5:
\begin{array}{l}
\left(2 y^2+3 y+\underline{\text{ }}\right)=2 \left(y^2+\frac{3 y}{2}+\underline{\text{ }}\right): \\
3 \left(x^2-\frac{4 x}{3}+\underline{\text{ }}\right)+\fbox{$2 \left(y^2+\frac{3 y}{2}+\underline{\text{ }}\right)$}=\underline{\text{ }}-4 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-4}{3}}{2}\right)^2=\frac{4}{9} \text{on }\text{the }\text{left }\text{and }3\times \frac{4}{9}=\frac{4}{3} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{4}{3}-4=-\frac{8}{3}: \\
3 \left(x^2-\frac{4 x}{3}+\frac{4}{9}\right)+2 \left(y^2+\frac{3 y}{2}+\underline{\text{ }}\right)=\fbox{$-\frac{8}{3}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{3}{2}}{2}\right)^2=\frac{9}{16} \text{on }\text{the }\text{left }\text{and }2\times \frac{9}{16}=\frac{9}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{9}{8}-\frac{8}{3}=-\frac{37}{24}: \\
3 \left(x^2-\frac{4 x}{3}+\frac{4}{9}\right)+2 \left(y^2+\frac{3 y}{2}+\frac{9}{16}\right)=\fbox{$-\frac{37}{24}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{4 x}{3}+\frac{4}{9}=\left(x-\frac{2}{3}\right)^2: \\
3 \fbox{$\left(x-\frac{2}{3}\right)^2$}+2 \left(y^2+\frac{3 y}{2}+\frac{9}{16}\right)=-\frac{37}{24} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{3 y}{2}+\frac{9}{16}=\left(y+\frac{3}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 3 \left(x-\frac{2}{3}\right)^2+2 \fbox{$\left(y+\frac{3}{4}\right)^2$}=-\frac{37}{24} \\
\end{array}
| khanacademy | amps |
Given the equation $-2 x^2+9 x-9 y-10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-2 x^2+9 x+(-9 y-10)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }-9 y-2 x^2+9 x-10 \text{from }\text{both }\text{sides}: \\
2 x^2-9 x+(9 y+10)=0 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Subtract }9 y+10 \text{from }\text{both }\text{sides}: \\
2 x^2-9 x=-9 y-10 \\
\end{array}
Step 4:
\begin{array}{l}
\text{Group }\text{terms }\text{together }\text{on }\text{the }\text{left }\text{hand }\text{side, }\text{with }\text{a }\text{placeholder }\text{constant}: \\
\left(2 x^2-9 x+\underline{\text{ }}\right)=(-9 y-10)+\underline{\text{ }} \\
\end{array}
Step 5:
\begin{array}{l}
\left(2 x^2-9 x+\underline{\text{ }}\right)=2 \left(x^2-\frac{9 x}{2}+\underline{\text{ }}\right): \\
\fbox{$2 \left(x^2-\frac{9 x}{2}+\underline{\text{ }}\right)$}=(-9 y-10)+\underline{\text{ }} \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-9}{2}}{2}\right)^2=\frac{81}{16} \text{on }\text{the }\text{left }\text{and }2\times \frac{81}{16}=\frac{81}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
(-9 y-10)+\frac{81}{8}=\frac{1}{8}-9 y: \\
2 \left(x^2-\frac{9 x}{2}+\frac{81}{16}\right)=\fbox{$\frac{1}{8}-9 y$} \\
\end{array}
Step 8:
\begin{array}{l}
x^2-\frac{9 x}{2}+\frac{81}{16}=\left(x-\frac{9}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 2 \fbox{$\left(x-\frac{9}{4}\right)^2$}=\frac{1}{8}-9 y \\
\end{array}
| khanacademy | amps |
Given the equation $-x^2+8 x-10 y^2-7 y-5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-10 y^2-7 y-x^2+8 x-5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }5 \text{to }\text{both }\text{sides}: \\
-10 y^2-7 y-x^2+8 x=5 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-x^2+8 x+\underline{\text{ }}\right)+\left(-10 y^2-7 y+\underline{\text{ }}\right)=\underline{\text{ }}+5 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-x^2+8 x+\underline{\text{ }}\right)=-\left(x^2-8 x+\underline{\text{ }}\right): \\
\fbox{$-\left(x^2-8 x+\underline{\text{ }}\right)$}+\left(-10 y^2-7 y+\underline{\text{ }}\right)=\underline{\text{ }}+5 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-10 y^2-7 y+\underline{\text{ }}\right)=-10 \left(y^2+\frac{7 y}{10}+\underline{\text{ }}\right): \\
-\left(x^2-8 x+\underline{\text{ }}\right)+\fbox{$-10 \left(y^2+\frac{7 y}{10}+\underline{\text{ }}\right)$}=\underline{\text{ }}+5 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-1 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-8}{2}\right)^2=16 \text{on }\text{the }\text{left }\text{and }-16=-16 \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
5-16=-11: \\
-\left(x^2-8 x+16\right)-10 \left(y^2+\frac{7 y}{10}+\underline{\text{ }}\right)=\fbox{$-11$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-10 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{7}{10}}{2}\right)^2=\frac{49}{400} \text{on }\text{the }\text{left }\text{and }-10\times \frac{49}{400}=-\frac{49}{40} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-11-\frac{49}{40}=-\frac{489}{40}: \\
-\left(x^2-8 x+16\right)-10 \left(y^2+\frac{7 y}{10}+\frac{49}{400}\right)=\fbox{$-\frac{489}{40}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-8 x+16=(x-4)^2: \\
-\fbox{$(x-4)^2$}-10 \left(y^2+\frac{7 y}{10}+\frac{49}{400}\right)=-\frac{489}{40} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{7 y}{10}+\frac{49}{400}=\left(y+\frac{7}{20}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -(x-4)^2-\text{10 }\fbox{$\left(y+\frac{7}{20}\right)^2$}=-\frac{489}{40} \\
\end{array}
| khanacademy | amps |
Given the equation $4 x^2+4 x+6 y^2-2 y-9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
6 y^2-2 y+4 x^2+4 x-9=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }9 \text{to }\text{both }\text{sides}: \\
6 y^2-2 y+4 x^2+4 x=9 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(4 x^2+4 x+\underline{\text{ }}\right)+\left(6 y^2-2 y+\underline{\text{ }}\right)=\underline{\text{ }}+9 \\
\end{array}
Step 4:
\begin{array}{l}
\left(4 x^2+4 x+\underline{\text{ }}\right)=4 \left(x^2+x+\underline{\text{ }}\right): \\
\fbox{$4 \left(x^2+x+\underline{\text{ }}\right)$}+\left(6 y^2-2 y+\underline{\text{ }}\right)=\underline{\text{ }}+9 \\
\end{array}
Step 5:
\begin{array}{l}
\left(6 y^2-2 y+\underline{\text{ }}\right)=6 \left(y^2-\frac{y}{3}+\underline{\text{ }}\right): \\
4 \left(x^2+x+\underline{\text{ }}\right)+\fbox{$6 \left(y^2-\frac{y}{3}+\underline{\text{ }}\right)$}=\underline{\text{ }}+9 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{4}{4}=1 \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
9+1=10: \\
4 \left(x^2+x+\frac{1}{4}\right)+6 \left(y^2-\frac{y}{3}+\underline{\text{ }}\right)=\fbox{$10$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-1}{3}}{2}\right)^2=\frac{1}{36} \text{on }\text{the }\text{left }\text{and }\frac{6}{36}=\frac{1}{6} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
10+\frac{1}{6}=\frac{61}{6}: \\
4 \left(x^2+x+\frac{1}{4}\right)+6 \left(y^2-\frac{y}{3}+\frac{1}{36}\right)=\fbox{$\frac{61}{6}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+x+\frac{1}{4}=\left(x+\frac{1}{2}\right)^2: \\
4 \fbox{$\left(x+\frac{1}{2}\right)^2$}+6 \left(y^2-\frac{y}{3}+\frac{1}{36}\right)=\frac{61}{6} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{y}{3}+\frac{1}{36}=\left(y-\frac{1}{6}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 4 \left(x+\frac{1}{2}\right)^2+6 \fbox{$\left(y-\frac{1}{6}\right)^2$}=\frac{61}{6} \\
\end{array}
| khanacademy | amps |
Given the equation $-3 x^2-6 x-2 y^2+8 y-10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-2 y^2+8 y-3 x^2-6 x-10=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }\text{10 }\text{to }\text{both }\text{sides}: \\
-2 y^2+8 y-3 x^2-6 x=10 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-3 x^2-6 x+\underline{\text{ }}\right)+\left(-2 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}+10 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-3 x^2-6 x+\underline{\text{ }}\right)=-3 \left(x^2+2 x+\underline{\text{ }}\right): \\
\fbox{$-3 \left(x^2+2 x+\underline{\text{ }}\right)$}+\left(-2 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}+10 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-2 y^2+8 y+\underline{\text{ }}\right)=-2 \left(y^2-4 y+\underline{\text{ }}\right): \\
-3 \left(x^2+2 x+\underline{\text{ }}\right)+\fbox{$-2 \left(y^2-4 y+\underline{\text{ }}\right)$}=\underline{\text{ }}+10 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{2}{2}\right)^2=1 \text{on }\text{the }\text{left }\text{and }-3\times 1=-3 \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
10-3=7: \\
-3 \left(x^2+2 x+1\right)-2 \left(y^2-4 y+\underline{\text{ }}\right)=\fbox{$7$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-4}{2}\right)^2=4 \text{on }\text{the }\text{left }\text{and }-2\times 4=-8 \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
7-8=-1: \\
-3 \left(x^2+2 x+1\right)-2 \left(y^2-4 y+4\right)=\fbox{$-1$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+2 x+1=(x+1)^2: \\
-3 \fbox{$(x+1)^2$}-2 \left(y^2-4 y+4\right)=-1 \\
\end{array}
Step 11:
\begin{array}{l}
y^2-4 y+4=(y-2)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -3 (x+1)^2-2 \fbox{$(y-2)^2$}=-1 \\
\end{array}
| khanacademy | amps |
Given the equation $-6 x^2-x-4 y^2+2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-6 x^2-x+\left(2-4 y^2\right)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }-4 y^2-6 x^2-x+2 \text{from }\text{both }\text{sides}: \\
6 x^2+x+\left(4 y^2-2\right)=0 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Add }2 \text{to }\text{both }\text{sides}: \\
4 y^2+6 x^2+x=2 \\
\end{array}
Step 4:
\begin{array}{l}
\text{Group }\text{terms }\text{together }\text{on }\text{the }\text{left }\text{hand }\text{side, }\text{with }\text{a }\text{placeholder }\text{constant}: \\
\left(6 x^2+x+\underline{\text{ }}\right)+4 y^2=\underline{\text{ }}+2 \\
\end{array}
Step 5:
\begin{array}{l}
\left(6 x^2+x+\underline{\text{ }}\right)=6 \left(x^2+\frac{x}{6}+\underline{\text{ }}\right): \\
\fbox{$6 \left(x^2+\frac{x}{6}+\underline{\text{ }}\right)$}+4 y^2=\underline{\text{ }}+2 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{6}}{2}\right)^2=\frac{1}{144} \text{on }\text{the }\text{left }\text{and }\frac{6}{144}=\frac{1}{24} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
2+\frac{1}{24}=\frac{49}{24}: \\
6 \left(x^2+\frac{x}{6}+\frac{1}{144}\right)+4 y^2=\fbox{$\frac{49}{24}$} \\
\end{array}
Step 8:
\begin{array}{l}
x^2+\frac{x}{6}+\frac{1}{144}=\left(x+\frac{1}{12}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 6 \fbox{$\left(x+\frac{1}{12}\right)^2$}+4 y^2=\frac{49}{24} \\
\end{array}
| khanacademy | amps |
Given the equation $6 x^2+4 x-4 y^2-4 y+10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-4 y^2-4 y+6 x^2+4 x+10=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }\text{10 }\text{from }\text{both }\text{sides}: \\
-4 y^2-4 y+6 x^2+4 x=-10 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(6 x^2+4 x+\underline{\text{ }}\right)+\left(-4 y^2-4 y+\underline{\text{ }}\right)=\underline{\text{ }}-10 \\
\end{array}
Step 4:
\begin{array}{l}
\left(6 x^2+4 x+\underline{\text{ }}\right)=6 \left(x^2+\frac{2 x}{3}+\underline{\text{ }}\right): \\
\fbox{$6 \left(x^2+\frac{2 x}{3}+\underline{\text{ }}\right)$}+\left(-4 y^2-4 y+\underline{\text{ }}\right)=\underline{\text{ }}-10 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-4 y^2-4 y+\underline{\text{ }}\right)=-4 \left(y^2+y+\underline{\text{ }}\right): \\
6 \left(x^2+\frac{2 x}{3}+\underline{\text{ }}\right)+\fbox{$-4 \left(y^2+y+\underline{\text{ }}\right)$}=\underline{\text{ }}-10 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{2}{3}}{2}\right)^2=\frac{1}{9} \text{on }\text{the }\text{left }\text{and }\frac{6}{9}=\frac{2}{3} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{2}{3}-10=-\frac{28}{3}: \\
6 \left(x^2+\frac{2 x}{3}+\frac{1}{9}\right)-4 \left(y^2+y+\underline{\text{ }}\right)=\fbox{$-\frac{28}{3}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{-4}{4}=-1 \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{28}{3}-1=-\frac{31}{3}: \\
6 \left(x^2+\frac{2 x}{3}+\frac{1}{9}\right)-4 \left(y^2+y+\frac{1}{4}\right)=\fbox{$-\frac{31}{3}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{2 x}{3}+\frac{1}{9}=\left(x+\frac{1}{3}\right)^2: \\
6 \fbox{$\left(x+\frac{1}{3}\right)^2$}-4 \left(y^2+y+\frac{1}{4}\right)=-\frac{31}{3} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+y+\frac{1}{4}=\left(y+\frac{1}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 6 \left(x+\frac{1}{3}\right)^2-4 \fbox{$\left(y+\frac{1}{2}\right)^2$}=-\frac{31}{3} \\
\end{array}
| khanacademy | amps |
Given the equation $4 x^2-9 x-3 y^2+7 y+9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-3 y^2+7 y+4 x^2-9 x+9=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }9 \text{from }\text{both }\text{sides}: \\
-3 y^2+7 y+4 x^2-9 x=-9 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(4 x^2-9 x+\underline{\text{ }}\right)+\left(-3 y^2+7 y+\underline{\text{ }}\right)=\underline{\text{ }}-9 \\
\end{array}
Step 4:
\begin{array}{l}
\left(4 x^2-9 x+\underline{\text{ }}\right)=4 \left(x^2-\frac{9 x}{4}+\underline{\text{ }}\right): \\
\fbox{$4 \left(x^2-\frac{9 x}{4}+\underline{\text{ }}\right)$}+\left(-3 y^2+7 y+\underline{\text{ }}\right)=\underline{\text{ }}-9 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-3 y^2+7 y+\underline{\text{ }}\right)=-3 \left(y^2-\frac{7 y}{3}+\underline{\text{ }}\right): \\
4 \left(x^2-\frac{9 x}{4}+\underline{\text{ }}\right)+\fbox{$-3 \left(y^2-\frac{7 y}{3}+\underline{\text{ }}\right)$}=\underline{\text{ }}-9 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-9}{4}}{2}\right)^2=\frac{81}{64} \text{on }\text{the }\text{left }\text{and }4\times \frac{81}{64}=\frac{81}{16} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{81}{16}-9=-\frac{63}{16}: \\
4 \left(x^2-\frac{9 x}{4}+\frac{81}{64}\right)-3 \left(y^2-\frac{7 y}{3}+\underline{\text{ }}\right)=\fbox{$-\frac{63}{16}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-7}{3}}{2}\right)^2=\frac{49}{36} \text{on }\text{the }\text{left }\text{and }-3\times \frac{49}{36}=-\frac{49}{12} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{63}{16}-\frac{49}{12}=-\frac{385}{48}: \\
4 \left(x^2-\frac{9 x}{4}+\frac{81}{64}\right)-3 \left(y^2-\frac{7 y}{3}+\frac{49}{36}\right)=\fbox{$-\frac{385}{48}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{9 x}{4}+\frac{81}{64}=\left(x-\frac{9}{8}\right)^2: \\
4 \fbox{$\left(x-\frac{9}{8}\right)^2$}-3 \left(y^2-\frac{7 y}{3}+\frac{49}{36}\right)=-\frac{385}{48} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{7 y}{3}+\frac{49}{36}=\left(y-\frac{7}{6}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 4 \left(x-\frac{9}{8}\right)^2-3 \fbox{$\left(y-\frac{7}{6}\right)^2$}=-\frac{385}{48} \\
\end{array}
| khanacademy | amps |
Given the equation $-8 x^2+9 x+5 y^2-4 y-3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
5 y^2-4 y-8 x^2+9 x-3=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }3 \text{to }\text{both }\text{sides}: \\
5 y^2-4 y-8 x^2+9 x=3 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-8 x^2+9 x+\underline{\text{ }}\right)+\left(5 y^2-4 y+\underline{\text{ }}\right)=\underline{\text{ }}+3 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-8 x^2+9 x+\underline{\text{ }}\right)=-8 \left(x^2-\frac{9 x}{8}+\underline{\text{ }}\right): \\
\fbox{$-8 \left(x^2-\frac{9 x}{8}+\underline{\text{ }}\right)$}+\left(5 y^2-4 y+\underline{\text{ }}\right)=\underline{\text{ }}+3 \\
\end{array}
Step 5:
\begin{array}{l}
\left(5 y^2-4 y+\underline{\text{ }}\right)=5 \left(y^2-\frac{4 y}{5}+\underline{\text{ }}\right): \\
-8 \left(x^2-\frac{9 x}{8}+\underline{\text{ }}\right)+\fbox{$5 \left(y^2-\frac{4 y}{5}+\underline{\text{ }}\right)$}=\underline{\text{ }}+3 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-9}{8}}{2}\right)^2=\frac{81}{256} \text{on }\text{the }\text{left }\text{and }-8\times \frac{81}{256}=-\frac{81}{32} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
3-\frac{81}{32}=\frac{15}{32}: \\
-8 \left(x^2-\frac{9 x}{8}+\frac{81}{256}\right)+5 \left(y^2-\frac{4 y}{5}+\underline{\text{ }}\right)=\fbox{$\frac{15}{32}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-4}{5}}{2}\right)^2=\frac{4}{25} \text{on }\text{the }\text{left }\text{and }5\times \frac{4}{25}=\frac{4}{5} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{15}{32}+\frac{4}{5}=\frac{203}{160}: \\
-8 \left(x^2-\frac{9 x}{8}+\frac{81}{256}\right)+5 \left(y^2-\frac{4 y}{5}+\frac{4}{25}\right)=\fbox{$\frac{203}{160}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{9 x}{8}+\frac{81}{256}=\left(x-\frac{9}{16}\right)^2: \\
-8 \fbox{$\left(x-\frac{9}{16}\right)^2$}+5 \left(y^2-\frac{4 y}{5}+\frac{4}{25}\right)=\frac{203}{160} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{4 y}{5}+\frac{4}{25}=\left(y-\frac{2}{5}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -8 \left(x-\frac{9}{16}\right)^2+5 \fbox{$\left(y-\frac{2}{5}\right)^2$}=\frac{203}{160} \\
\end{array}
| khanacademy | amps |
Given the equation $-7 x^2-x+7 y^2-8=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-7 x^2-x+\left(7 y^2-8\right)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }7 y^2-7 x^2-x-8 \text{from }\text{both }\text{sides}: \\
7 x^2+x+\left(8-7 y^2\right)=0 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Subtract }8 \text{from }\text{both }\text{sides}: \\
-7 y^2+7 x^2+x=-8 \\
\end{array}
Step 4:
\begin{array}{l}
\text{Group }\text{terms }\text{together }\text{on }\text{the }\text{left }\text{hand }\text{side, }\text{with }\text{a }\text{placeholder }\text{constant}: \\
\left(7 x^2+x+\underline{\text{ }}\right)-7 y^2=\underline{\text{ }}-8 \\
\end{array}
Step 5:
\begin{array}{l}
\left(7 x^2+x+\underline{\text{ }}\right)=7 \left(x^2+\frac{x}{7}+\underline{\text{ }}\right): \\
\fbox{$7 \left(x^2+\frac{x}{7}+\underline{\text{ }}\right)$}-7 y^2=\underline{\text{ }}-8 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{7}}{2}\right)^2=\frac{1}{196} \text{on }\text{the }\text{left }\text{and }\frac{7}{196}=\frac{1}{28} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{1}{28}-8=-\frac{223}{28}: \\
7 \left(x^2+\frac{x}{7}+\frac{1}{196}\right)-7 y^2=\fbox{$-\frac{223}{28}$} \\
\end{array}
Step 8:
\begin{array}{l}
x^2+\frac{x}{7}+\frac{1}{196}=\left(x+\frac{1}{14}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 7 \fbox{$\left(x+\frac{1}{14}\right)^2$}-7 y^2=-\frac{223}{28} \\
\end{array}
| khanacademy | amps |
Given the equation $-10 x^2-x+9 y^2-3 y+3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
9 y^2-3 y-10 x^2-x+3=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }3 \text{from }\text{both }\text{sides}: \\
9 y^2-3 y-10 x^2-x=-3 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-10 x^2-x+\underline{\text{ }}\right)+\left(9 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}-3 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-10 x^2-x+\underline{\text{ }}\right)=-10 \left(x^2+\frac{x}{10}+\underline{\text{ }}\right): \\
\fbox{$-10 \left(x^2+\frac{x}{10}+\underline{\text{ }}\right)$}+\left(9 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}-3 \\
\end{array}
Step 5:
\begin{array}{l}
\left(9 y^2-3 y+\underline{\text{ }}\right)=9 \left(y^2-\frac{y}{3}+\underline{\text{ }}\right): \\
-10 \left(x^2+\frac{x}{10}+\underline{\text{ }}\right)+\fbox{$9 \left(y^2-\frac{y}{3}+\underline{\text{ }}\right)$}=\underline{\text{ }}-3 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-10 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{10}}{2}\right)^2=\frac{1}{400} \text{on }\text{the }\text{left }\text{and }\frac{-10}{400}=-\frac{1}{40} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-3-\frac{1}{40}=-\frac{121}{40}: \\
-10 \left(x^2+\frac{x}{10}+\frac{1}{400}\right)+9 \left(y^2-\frac{y}{3}+\underline{\text{ }}\right)=\fbox{$-\frac{121}{40}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-1}{3}}{2}\right)^2=\frac{1}{36} \text{on }\text{the }\text{left }\text{and }\frac{9}{36}=\frac{1}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{1}{4}-\frac{121}{40}=-\frac{111}{40}: \\
-10 \left(x^2+\frac{x}{10}+\frac{1}{400}\right)+9 \left(y^2-\frac{y}{3}+\frac{1}{36}\right)=\fbox{$-\frac{111}{40}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{x}{10}+\frac{1}{400}=\left(x+\frac{1}{20}\right)^2: \\
-10 \fbox{$\left(x+\frac{1}{20}\right)^2$}+9 \left(y^2-\frac{y}{3}+\frac{1}{36}\right)=-\frac{111}{40} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{y}{3}+\frac{1}{36}=\left(y-\frac{1}{6}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -10 \left(x+\frac{1}{20}\right)^2+9 \fbox{$\left(y-\frac{1}{6}\right)^2$}=-\frac{111}{40} \\
\end{array}
| khanacademy | amps |
Given the equation $-2 x^2-7 x-6 y^2-9 y+10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-6 y^2-9 y-2 x^2-7 x+10=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }\text{10 }\text{from }\text{both }\text{sides}: \\
-6 y^2-9 y-2 x^2-7 x=-10 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-2 x^2-7 x+\underline{\text{ }}\right)+\left(-6 y^2-9 y+\underline{\text{ }}\right)=\underline{\text{ }}-10 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-2 x^2-7 x+\underline{\text{ }}\right)=-2 \left(x^2+\frac{7 x}{2}+\underline{\text{ }}\right): \\
\fbox{$-2 \left(x^2+\frac{7 x}{2}+\underline{\text{ }}\right)$}+\left(-6 y^2-9 y+\underline{\text{ }}\right)=\underline{\text{ }}-10 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-6 y^2-9 y+\underline{\text{ }}\right)=-6 \left(y^2+\frac{3 y}{2}+\underline{\text{ }}\right): \\
-2 \left(x^2+\frac{7 x}{2}+\underline{\text{ }}\right)+\fbox{$-6 \left(y^2+\frac{3 y}{2}+\underline{\text{ }}\right)$}=\underline{\text{ }}-10 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{7}{2}}{2}\right)^2=\frac{49}{16} \text{on }\text{the }\text{left }\text{and }-2\times \frac{49}{16}=-\frac{49}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-10-\frac{49}{8}=-\frac{129}{8}: \\
-2 \left(x^2+\frac{7 x}{2}+\frac{49}{16}\right)-6 \left(y^2+\frac{3 y}{2}+\underline{\text{ }}\right)=\fbox{$-\frac{129}{8}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{3}{2}}{2}\right)^2=\frac{9}{16} \text{on }\text{the }\text{left }\text{and }-6\times \frac{9}{16}=-\frac{27}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{129}{8}-\frac{27}{8}=-\frac{39}{2}: \\
-2 \left(x^2+\frac{7 x}{2}+\frac{49}{16}\right)-6 \left(y^2+\frac{3 y}{2}+\frac{9}{16}\right)=\fbox{$-\frac{39}{2}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{7 x}{2}+\frac{49}{16}=\left(x+\frac{7}{4}\right)^2: \\
-2 \fbox{$\left(x+\frac{7}{4}\right)^2$}-6 \left(y^2+\frac{3 y}{2}+\frac{9}{16}\right)=-\frac{39}{2} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{3 y}{2}+\frac{9}{16}=\left(y+\frac{3}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -2 \left(x+\frac{7}{4}\right)^2-6 \fbox{$\left(y+\frac{3}{4}\right)^2$}=-\frac{39}{2} \\
\end{array}
| khanacademy | amps |
Given the equation $5 x^2+6 x+7 y^2+8 y+3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
7 y^2+8 y+5 x^2+6 x+3=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }3 \text{from }\text{both }\text{sides}: \\
7 y^2+8 y+5 x^2+6 x=-3 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(5 x^2+6 x+\underline{\text{ }}\right)+\left(7 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}-3 \\
\end{array}
Step 4:
\begin{array}{l}
\left(5 x^2+6 x+\underline{\text{ }}\right)=5 \left(x^2+\frac{6 x}{5}+\underline{\text{ }}\right): \\
\fbox{$5 \left(x^2+\frac{6 x}{5}+\underline{\text{ }}\right)$}+\left(7 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}-3 \\
\end{array}
Step 5:
\begin{array}{l}
\left(7 y^2+8 y+\underline{\text{ }}\right)=7 \left(y^2+\frac{8 y}{7}+\underline{\text{ }}\right): \\
5 \left(x^2+\frac{6 x}{5}+\underline{\text{ }}\right)+\fbox{$7 \left(y^2+\frac{8 y}{7}+\underline{\text{ }}\right)$}=\underline{\text{ }}-3 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{6}{5}}{2}\right)^2=\frac{9}{25} \text{on }\text{the }\text{left }\text{and }5\times \frac{9}{25}=\frac{9}{5} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{9}{5}-3=-\frac{6}{5}: \\
5 \left(x^2+\frac{6 x}{5}+\frac{9}{25}\right)+7 \left(y^2+\frac{8 y}{7}+\underline{\text{ }}\right)=\fbox{$-\frac{6}{5}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{8}{7}}{2}\right)^2=\frac{16}{49} \text{on }\text{the }\text{left }\text{and }7\times \frac{16}{49}=\frac{16}{7} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{16}{7}-\frac{6}{5}=\frac{38}{35}: \\
5 \left(x^2+\frac{6 x}{5}+\frac{9}{25}\right)+7 \left(y^2+\frac{8 y}{7}+\frac{16}{49}\right)=\fbox{$\frac{38}{35}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{6 x}{5}+\frac{9}{25}=\left(x+\frac{3}{5}\right)^2: \\
5 \fbox{$\left(x+\frac{3}{5}\right)^2$}+7 \left(y^2+\frac{8 y}{7}+\frac{16}{49}\right)=\frac{38}{35} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{8 y}{7}+\frac{16}{49}=\left(y+\frac{4}{7}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 5 \left(x+\frac{3}{5}\right)^2+7 \fbox{$\left(y+\frac{4}{7}\right)^2$}=\frac{38}{35} \\
\end{array}
| khanacademy | amps |
Given the equation $-5 x^2-7 x+10 y-1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-5 x^2-7 x+(10 y-1)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }10 y-5 x^2-7 x-1 \text{from }\text{both }\text{sides}: \\
5 x^2+7 x+(1-10 y)=0 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Subtract }1-10 y \text{from }\text{both }\text{sides}: \\
5 x^2+7 x=10 y-1 \\
\end{array}
Step 4:
\begin{array}{l}
\text{Group }\text{terms }\text{together }\text{on }\text{the }\text{left }\text{hand }\text{side, }\text{with }\text{a }\text{placeholder }\text{constant}: \\
\left(5 x^2+7 x+\underline{\text{ }}\right)=(10 y-1)+\underline{\text{ }} \\
\end{array}
Step 5:
\begin{array}{l}
\left(5 x^2+7 x+\underline{\text{ }}\right)=5 \left(x^2+\frac{7 x}{5}+\underline{\text{ }}\right): \\
\fbox{$5 \left(x^2+\frac{7 x}{5}+\underline{\text{ }}\right)$}=(10 y-1)+\underline{\text{ }} \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{7}{5}}{2}\right)^2=\frac{49}{100} \text{on }\text{the }\text{left }\text{and }5\times \frac{49}{100}=\frac{49}{20} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
(10 y-1)+\frac{49}{20}=10 y+\frac{29}{20}: \\
5 \left(x^2+\frac{7 x}{5}+\frac{49}{100}\right)=\fbox{$10 y+\frac{29}{20}$} \\
\end{array}
Step 8:
\begin{array}{l}
x^2+\frac{7 x}{5}+\frac{49}{100}=\left(x+\frac{7}{10}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 5 \fbox{$\left(x+\frac{7}{10}\right)^2$}=10 y+\frac{29}{20} \\
\end{array}
| khanacademy | amps |
Given the equation $-6 x^2-2 x+6 y^2+8 y+3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
6 y^2+8 y-6 x^2-2 x+3=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }3 \text{from }\text{both }\text{sides}: \\
6 y^2+8 y-6 x^2-2 x=-3 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-6 x^2-2 x+\underline{\text{ }}\right)+\left(6 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}-3 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-6 x^2-2 x+\underline{\text{ }}\right)=-6 \left(x^2+\frac{x}{3}+\underline{\text{ }}\right): \\
\fbox{$-6 \left(x^2+\frac{x}{3}+\underline{\text{ }}\right)$}+\left(6 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}-3 \\
\end{array}
Step 5:
\begin{array}{l}
\left(6 y^2+8 y+\underline{\text{ }}\right)=6 \left(y^2+\frac{4 y}{3}+\underline{\text{ }}\right): \\
-6 \left(x^2+\frac{x}{3}+\underline{\text{ }}\right)+\fbox{$6 \left(y^2+\frac{4 y}{3}+\underline{\text{ }}\right)$}=\underline{\text{ }}-3 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{3}}{2}\right)^2=\frac{1}{36} \text{on }\text{the }\text{left }\text{and }\frac{-6}{36}=-\frac{1}{6} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-3-\frac{1}{6}=-\frac{19}{6}: \\
-6 \left(x^2+\frac{x}{3}+\frac{1}{36}\right)+6 \left(y^2+\frac{4 y}{3}+\underline{\text{ }}\right)=\fbox{$-\frac{19}{6}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{4}{3}}{2}\right)^2=\frac{4}{9} \text{on }\text{the }\text{left }\text{and }6\times \frac{4}{9}=\frac{8}{3} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{8}{3}-\frac{19}{6}=-\frac{1}{2}: \\
-6 \left(x^2+\frac{x}{3}+\frac{1}{36}\right)+6 \left(y^2+\frac{4 y}{3}+\frac{4}{9}\right)=\fbox{$-\frac{1}{2}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{x}{3}+\frac{1}{36}=\left(x+\frac{1}{6}\right)^2: \\
-6 \fbox{$\left(x+\frac{1}{6}\right)^2$}+6 \left(y^2+\frac{4 y}{3}+\frac{4}{9}\right)=-\frac{1}{2} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{4 y}{3}+\frac{4}{9}=\left(y+\frac{2}{3}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -6 \left(x+\frac{1}{6}\right)^2+6 \fbox{$\left(y+\frac{2}{3}\right)^2$}=-\frac{1}{2} \\
\end{array}
| khanacademy | amps |
Given the equation $6 x^2-10 x+8 y^2-2 y-2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
8 y^2-2 y+6 x^2-10 x-2=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }2 \text{to }\text{both }\text{sides}: \\
8 y^2-2 y+6 x^2-10 x=2 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(6 x^2-10 x+\underline{\text{ }}\right)+\left(8 y^2-2 y+\underline{\text{ }}\right)=\underline{\text{ }}+2 \\
\end{array}
Step 4:
\begin{array}{l}
\left(6 x^2-10 x+\underline{\text{ }}\right)=6 \left(x^2-\frac{5 x}{3}+\underline{\text{ }}\right): \\
\fbox{$6 \left(x^2-\frac{5 x}{3}+\underline{\text{ }}\right)$}+\left(8 y^2-2 y+\underline{\text{ }}\right)=\underline{\text{ }}+2 \\
\end{array}
Step 5:
\begin{array}{l}
\left(8 y^2-2 y+\underline{\text{ }}\right)=8 \left(y^2-\frac{y}{4}+\underline{\text{ }}\right): \\
6 \left(x^2-\frac{5 x}{3}+\underline{\text{ }}\right)+\fbox{$8 \left(y^2-\frac{y}{4}+\underline{\text{ }}\right)$}=\underline{\text{ }}+2 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-5}{3}}{2}\right)^2=\frac{25}{36} \text{on }\text{the }\text{left }\text{and }6\times \frac{25}{36}=\frac{25}{6} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
2+\frac{25}{6}=\frac{37}{6}: \\
6 \left(x^2-\frac{5 x}{3}+\frac{25}{36}\right)+8 \left(y^2-\frac{y}{4}+\underline{\text{ }}\right)=\fbox{$\frac{37}{6}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-1}{4}}{2}\right)^2=\frac{1}{64} \text{on }\text{the }\text{left }\text{and }\frac{8}{64}=\frac{1}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{37}{6}+\frac{1}{8}=\frac{151}{24}: \\
6 \left(x^2-\frac{5 x}{3}+\frac{25}{36}\right)+8 \left(y^2-\frac{y}{4}+\frac{1}{64}\right)=\fbox{$\frac{151}{24}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{5 x}{3}+\frac{25}{36}=\left(x-\frac{5}{6}\right)^2: \\
6 \fbox{$\left(x-\frac{5}{6}\right)^2$}+8 \left(y^2-\frac{y}{4}+\frac{1}{64}\right)=\frac{151}{24} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{y}{4}+\frac{1}{64}=\left(y-\frac{1}{8}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 6 \left(x-\frac{5}{6}\right)^2+8 \fbox{$\left(y-\frac{1}{8}\right)^2$}=\frac{151}{24} \\
\end{array}
| khanacademy | amps |
Given the equation $x^2-7 x+y^2+4 y-10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
y^2+4 y+x^2-7 x-10=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }\text{10 }\text{to }\text{both }\text{sides}: \\
y^2+4 y+x^2-7 x=10 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(x^2-7 x+\underline{\text{ }}\right)+\left(y^2+4 y+\underline{\text{ }}\right)=\underline{\text{ }}+10 \\
\end{array}
Step 4:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it, }\text{then }\text{add }\text{it }\text{to }\text{both }\text{sides.} \\
\text{Add }\left(\frac{-7}{2}\right)^2=\frac{49}{4} \text{to }\text{both }\text{sides}: \\
\end{array}
Step 5:
\begin{array}{l}
10+\frac{49}{4}=\frac{89}{4}: \\
\left(x^2-7 x+\frac{49}{4}\right)+\left(y^2+4 y+\underline{\text{ }}\right)=\fbox{$\frac{89}{4}$} \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it, }\text{then }\text{add }\text{it }\text{to }\text{both }\text{sides.} \\
\text{Add }\left(\frac{4}{2}\right)^2=4 \text{to }\text{both }\text{sides}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{89}{4}+4=\frac{105}{4}: \\
\left(x^2-7 x+\frac{49}{4}\right)+\left(y^2+4 y+4\right)=\fbox{$\frac{105}{4}$} \\
\end{array}
Step 8:
\begin{array}{l}
x^2-7 x+\frac{49}{4}=\left(x-\frac{7}{2}\right)^2: \\
\fbox{$\left(x-\frac{7}{2}\right)^2$}+\left(y^2+4 y+4\right)=\frac{105}{4} \\
\end{array}
Step 9:
\begin{array}{l}
y^2+4 y+4=(y+2)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & \left(x-\frac{7}{2}\right)^2+\fbox{$(y+2)^2$}=\frac{105}{4} \\
\end{array}
| khanacademy | amps |
Given the equation $9 x^2+x+2 y+6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
9 x^2+x+(2 y+6)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }2 y+6 \text{from }\text{both }\text{sides}: \\
9 x^2+x=-2 y-6 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{together }\text{on }\text{the }\text{left }\text{hand }\text{side, }\text{with }\text{a }\text{placeholder }\text{constant}: \\
\left(9 x^2+x+\underline{\text{ }}\right)=(-2 y-6)+\underline{\text{ }} \\
\end{array}
Step 4:
\begin{array}{l}
\left(9 x^2+x+\underline{\text{ }}\right)=9 \left(x^2+\frac{x}{9}+\underline{\text{ }}\right): \\
\fbox{$9 \left(x^2+\frac{x}{9}+\underline{\text{ }}\right)$}=(-2 y-6)+\underline{\text{ }} \\
\end{array}
Step 5:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{9}}{2}\right)^2=\frac{1}{324} \text{on }\text{the }\text{left }\text{and }\frac{9}{324}=\frac{1}{36} \text{on }\text{the }\text{right}: \\
\end{array}
Step 6:
\begin{array}{l}
(-2 y-6)+\frac{1}{36}=-2 y-\frac{215}{36}: \\
9 \left(x^2+\frac{x}{9}+\frac{1}{324}\right)=\fbox{$-2 y-\frac{215}{36}$} \\
\end{array}
Step 7:
\begin{array}{l}
x^2+\frac{x}{9}+\frac{1}{324}=\left(x+\frac{1}{18}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 9 \fbox{$\left(x+\frac{1}{18}\right)^2$}=-2 y-\frac{215}{36} \\
\end{array}
| khanacademy | amps |
Given the equation $4 x^2-4 x-5 y^2+4 y-2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-5 y^2+4 y+4 x^2-4 x-2=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }2 \text{to }\text{both }\text{sides}: \\
-5 y^2+4 y+4 x^2-4 x=2 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(4 x^2-4 x+\underline{\text{ }}\right)+\left(-5 y^2+4 y+\underline{\text{ }}\right)=\underline{\text{ }}+2 \\
\end{array}
Step 4:
\begin{array}{l}
\left(4 x^2-4 x+\underline{\text{ }}\right)=4 \left(x^2-x+\underline{\text{ }}\right): \\
\fbox{$4 \left(x^2-x+\underline{\text{ }}\right)$}+\left(-5 y^2+4 y+\underline{\text{ }}\right)=\underline{\text{ }}+2 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-5 y^2+4 y+\underline{\text{ }}\right)=-5 \left(y^2-\frac{4 y}{5}+\underline{\text{ }}\right): \\
4 \left(x^2-x+\underline{\text{ }}\right)+\fbox{$-5 \left(y^2-\frac{4 y}{5}+\underline{\text{ }}\right)$}=\underline{\text{ }}+2 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{4}{4}=1 \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
2+1=3: \\
4 \left(x^2-x+\frac{1}{4}\right)-5 \left(y^2-\frac{4 y}{5}+\underline{\text{ }}\right)=\fbox{$3$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-4}{5}}{2}\right)^2=\frac{4}{25} \text{on }\text{the }\text{left }\text{and }-5\times \frac{4}{25}=-\frac{4}{5} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
3-\frac{4}{5}=\frac{11}{5}: \\
4 \left(x^2-x+\frac{1}{4}\right)-5 \left(y^2-\frac{4 y}{5}+\frac{4}{25}\right)=\fbox{$\frac{11}{5}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-x+\frac{1}{4}=\left(x-\frac{1}{2}\right)^2: \\
4 \fbox{$\left(x-\frac{1}{2}\right)^2$}-5 \left(y^2-\frac{4 y}{5}+\frac{4}{25}\right)=\frac{11}{5} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{4 y}{5}+\frac{4}{25}=\left(y-\frac{2}{5}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 4 \left(x-\frac{1}{2}\right)^2-5 \fbox{$\left(y-\frac{2}{5}\right)^2$}=\frac{11}{5} \\
\end{array}
| khanacademy | amps |
Given the equation $6 x^2-5 x+2 y^2-2 y-8=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
2 y^2-2 y+6 x^2-5 x-8=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }8 \text{to }\text{both }\text{sides}: \\
2 y^2-2 y+6 x^2-5 x=8 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(6 x^2-5 x+\underline{\text{ }}\right)+\left(2 y^2-2 y+\underline{\text{ }}\right)=\underline{\text{ }}+8 \\
\end{array}
Step 4:
\begin{array}{l}
\left(6 x^2-5 x+\underline{\text{ }}\right)=6 \left(x^2-\frac{5 x}{6}+\underline{\text{ }}\right): \\
\fbox{$6 \left(x^2-\frac{5 x}{6}+\underline{\text{ }}\right)$}+\left(2 y^2-2 y+\underline{\text{ }}\right)=\underline{\text{ }}+8 \\
\end{array}
Step 5:
\begin{array}{l}
\left(2 y^2-2 y+\underline{\text{ }}\right)=2 \left(y^2-y+\underline{\text{ }}\right): \\
6 \left(x^2-\frac{5 x}{6}+\underline{\text{ }}\right)+\fbox{$2 \left(y^2-y+\underline{\text{ }}\right)$}=\underline{\text{ }}+8 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-5}{6}}{2}\right)^2=\frac{25}{144} \text{on }\text{the }\text{left }\text{and }6\times \frac{25}{144}=\frac{25}{24} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
8+\frac{25}{24}=\frac{217}{24}: \\
6 \left(x^2-\frac{5 x}{6}+\frac{25}{144}\right)+2 \left(y^2-y+\underline{\text{ }}\right)=\fbox{$\frac{217}{24}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{2}{4}=\frac{1}{2} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{217}{24}+\frac{1}{2}=\frac{229}{24}: \\
6 \left(x^2-\frac{5 x}{6}+\frac{25}{144}\right)+2 \left(y^2-y+\frac{1}{4}\right)=\fbox{$\frac{229}{24}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{5 x}{6}+\frac{25}{144}=\left(x-\frac{5}{12}\right)^2: \\
6 \fbox{$\left(x-\frac{5}{12}\right)^2$}+2 \left(y^2-y+\frac{1}{4}\right)=\frac{229}{24} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-y+\frac{1}{4}=\left(y-\frac{1}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 6 \left(x-\frac{5}{12}\right)^2+2 \fbox{$\left(y-\frac{1}{2}\right)^2$}=\frac{229}{24} \\
\end{array}
| khanacademy | amps |
Given the equation $9 x^2-4 x+5 y^2-3 y+5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
5 y^2-3 y+9 x^2-4 x+5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }5 \text{from }\text{both }\text{sides}: \\
5 y^2-3 y+9 x^2-4 x=-5 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(9 x^2-4 x+\underline{\text{ }}\right)+\left(5 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\end{array}
Step 4:
\begin{array}{l}
\left(9 x^2-4 x+\underline{\text{ }}\right)=9 \left(x^2-\frac{4 x}{9}+\underline{\text{ }}\right): \\
\fbox{$9 \left(x^2-\frac{4 x}{9}+\underline{\text{ }}\right)$}+\left(5 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\end{array}
Step 5:
\begin{array}{l}
\left(5 y^2-3 y+\underline{\text{ }}\right)=5 \left(y^2-\frac{3 y}{5}+\underline{\text{ }}\right): \\
9 \left(x^2-\frac{4 x}{9}+\underline{\text{ }}\right)+\fbox{$5 \left(y^2-\frac{3 y}{5}+\underline{\text{ }}\right)$}=\underline{\text{ }}-5 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-4}{9}}{2}\right)^2=\frac{4}{81} \text{on }\text{the }\text{left }\text{and }9\times \frac{4}{81}=\frac{4}{9} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{4}{9}-5=-\frac{41}{9}: \\
9 \left(x^2-\frac{4 x}{9}+\frac{4}{81}\right)+5 \left(y^2-\frac{3 y}{5}+\underline{\text{ }}\right)=\fbox{$-\frac{41}{9}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-3}{5}}{2}\right)^2=\frac{9}{100} \text{on }\text{the }\text{left }\text{and }5\times \frac{9}{100}=\frac{9}{20} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{9}{20}-\frac{41}{9}=-\frac{739}{180}: \\
9 \left(x^2-\frac{4 x}{9}+\frac{4}{81}\right)+5 \left(y^2-\frac{3 y}{5}+\frac{9}{100}\right)=\fbox{$-\frac{739}{180}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{4 x}{9}+\frac{4}{81}=\left(x-\frac{2}{9}\right)^2: \\
9 \fbox{$\left(x-\frac{2}{9}\right)^2$}+5 \left(y^2-\frac{3 y}{5}+\frac{9}{100}\right)=-\frac{739}{180} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{3 y}{5}+\frac{9}{100}=\left(y-\frac{3}{10}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 9 \left(x-\frac{2}{9}\right)^2+5 \fbox{$\left(y-\frac{3}{10}\right)^2$}=-\frac{739}{180} \\
\end{array}
| khanacademy | amps |
Given the equation $-4 x^2-9 x+5 y^2-4 y=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
5 y^2-4 y-4 x^2-9 x=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-4 x^2-9 x+\underline{\text{ }}\right)+\left(5 y^2-4 y+\underline{\text{ }}\right)=\underline{\text{ }}+0 \\
\end{array}
Step 3:
\begin{array}{l}
\left(-4 x^2-9 x+\underline{\text{ }}\right)=-4 \left(x^2+\frac{9 x}{4}+\underline{\text{ }}\right): \\
\fbox{$-4 \left(x^2+\frac{9 x}{4}+\underline{\text{ }}\right)$}+\left(5 y^2-4 y+\underline{\text{ }}\right)=\underline{\text{ }}+0 \\
\end{array}
Step 4:
\begin{array}{l}
\left(5 y^2-4 y+\underline{\text{ }}\right)=5 \left(y^2-\frac{4 y}{5}+\underline{\text{ }}\right): \\
-4 \left(x^2+\frac{9 x}{4}+\underline{\text{ }}\right)+\fbox{$5 \left(y^2-\frac{4 y}{5}+\underline{\text{ }}\right)$}=\underline{\text{ }}+0 \\
\end{array}
Step 5:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{9}{4}}{2}\right)^2=\frac{81}{64} \text{on }\text{the }\text{left }\text{and }-4\times \frac{81}{64}=-\frac{81}{16} \text{on }\text{the }\text{right}: \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-4}{5}}{2}\right)^2=\frac{4}{25} \text{on }\text{the }\text{left }\text{and }5\times \frac{4}{25}=\frac{4}{5} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{4}{5}-\frac{81}{16}=-\frac{341}{80}: \\
-4 \left(x^2+\frac{9 x}{4}+\frac{81}{64}\right)+5 \left(y^2-\frac{4 y}{5}+\frac{4}{25}\right)=\fbox{$-\frac{341}{80}$} \\
\end{array}
Step 8:
\begin{array}{l}
x^2+\frac{9 x}{4}+\frac{81}{64}=\left(x+\frac{9}{8}\right)^2: \\
-4 \fbox{$\left(x+\frac{9}{8}\right)^2$}+5 \left(y^2-\frac{4 y}{5}+\frac{4}{25}\right)=-\frac{341}{80} \\
\end{array}
Step 9:
\begin{array}{l}
y^2-\frac{4 y}{5}+\frac{4}{25}=\left(y-\frac{2}{5}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -4 \left(x+\frac{9}{8}\right)^2+5 \fbox{$\left(y-\frac{2}{5}\right)^2$}=-\frac{341}{80} \\
\end{array}
| khanacademy | amps |
Given the equation $-7 x^2-4 x+5 y^2+y+9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
5 y^2+y-7 x^2-4 x+9=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }9 \text{from }\text{both }\text{sides}: \\
5 y^2+y-7 x^2-4 x=-9 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-7 x^2-4 x+\underline{\text{ }}\right)+\left(5 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}-9 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-7 x^2-4 x+\underline{\text{ }}\right)=-7 \left(x^2+\frac{4 x}{7}+\underline{\text{ }}\right): \\
\fbox{$-7 \left(x^2+\frac{4 x}{7}+\underline{\text{ }}\right)$}+\left(5 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}-9 \\
\end{array}
Step 5:
\begin{array}{l}
\left(5 y^2+y+\underline{\text{ }}\right)=5 \left(y^2+\frac{y}{5}+\underline{\text{ }}\right): \\
-7 \left(x^2+\frac{4 x}{7}+\underline{\text{ }}\right)+\fbox{$5 \left(y^2+\frac{y}{5}+\underline{\text{ }}\right)$}=\underline{\text{ }}-9 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{4}{7}}{2}\right)^2=\frac{4}{49} \text{on }\text{the }\text{left }\text{and }-7\times \frac{4}{49}=-\frac{4}{7} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-9-\frac{4}{7}=-\frac{67}{7}: \\
-7 \left(x^2+\frac{4 x}{7}+\frac{4}{49}\right)+5 \left(y^2+\frac{y}{5}+\underline{\text{ }}\right)=\fbox{$-\frac{67}{7}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{5}}{2}\right)^2=\frac{1}{100} \text{on }\text{the }\text{left }\text{and }\frac{5}{100}=\frac{1}{20} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{1}{20}-\frac{67}{7}=-\frac{1333}{140}: \\
-7 \left(x^2+\frac{4 x}{7}+\frac{4}{49}\right)+5 \left(y^2+\frac{y}{5}+\frac{1}{100}\right)=\fbox{$-\frac{1333}{140}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{4 x}{7}+\frac{4}{49}=\left(x+\frac{2}{7}\right)^2: \\
-7 \fbox{$\left(x+\frac{2}{7}\right)^2$}+5 \left(y^2+\frac{y}{5}+\frac{1}{100}\right)=-\frac{1333}{140} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{y}{5}+\frac{1}{100}=\left(y+\frac{1}{10}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -7 \left(x+\frac{2}{7}\right)^2+5 \fbox{$\left(y+\frac{1}{10}\right)^2$}=-\frac{1333}{140} \\
\end{array}
| khanacademy | amps |
Given the equation $-3 x^2+6 x-9 y^2-y+10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-9 y^2-y-3 x^2+6 x+10=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }\text{10 }\text{from }\text{both }\text{sides}: \\
-9 y^2-y-3 x^2+6 x=-10 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-3 x^2+6 x+\underline{\text{ }}\right)+\left(-9 y^2-y+\underline{\text{ }}\right)=\underline{\text{ }}-10 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-3 x^2+6 x+\underline{\text{ }}\right)=-3 \left(x^2-2 x+\underline{\text{ }}\right): \\
\fbox{$-3 \left(x^2-2 x+\underline{\text{ }}\right)$}+\left(-9 y^2-y+\underline{\text{ }}\right)=\underline{\text{ }}-10 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-9 y^2-y+\underline{\text{ }}\right)=-9 \left(y^2+\frac{y}{9}+\underline{\text{ }}\right): \\
-3 \left(x^2-2 x+\underline{\text{ }}\right)+\fbox{$-9 \left(y^2+\frac{y}{9}+\underline{\text{ }}\right)$}=\underline{\text{ }}-10 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-2}{2}\right)^2=1 \text{on }\text{the }\text{left }\text{and }-3\times 1=-3 \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-10-3=-13: \\
-3 \left(x^2-2 x+1\right)-9 \left(y^2+\frac{y}{9}+\underline{\text{ }}\right)=\fbox{$-13$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{9}}{2}\right)^2=\frac{1}{324} \text{on }\text{the }\text{left }\text{and }\frac{-9}{324}=-\frac{1}{36} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-13-\frac{1}{36}=-\frac{469}{36}: \\
-3 \left(x^2-2 x+1\right)-9 \left(y^2+\frac{y}{9}+\frac{1}{324}\right)=\fbox{$-\frac{469}{36}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-2 x+1=(x-1)^2: \\
-3 \fbox{$(x-1)^2$}-9 \left(y^2+\frac{y}{9}+\frac{1}{324}\right)=-\frac{469}{36} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{y}{9}+\frac{1}{324}=\left(y+\frac{1}{18}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -3 (x-1)^2-9 \fbox{$\left(y+\frac{1}{18}\right)^2$}=-\frac{469}{36} \\
\end{array}
| khanacademy | amps |
Given the equation $8 x^2+10 x+3 y=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
3 y+8 x^2+10 x=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }3 y \text{from }\text{both }\text{sides}: \\
8 x^2+10 x=-3 y \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{together }\text{on }\text{the }\text{left }\text{hand }\text{side, }\text{with }\text{a }\text{placeholder }\text{constant}: \\
\left(8 x^2+10 x+\underline{\text{ }}\right)=\underline{\text{ }}-3 y \\
\end{array}
Step 4:
\begin{array}{l}
\left(8 x^2+10 x+\underline{\text{ }}\right)=8 \left(x^2+\frac{5 x}{4}+\underline{\text{ }}\right): \\
\fbox{$8 \left(x^2+\frac{5 x}{4}+\underline{\text{ }}\right)$}=\underline{\text{ }}-3 y \\
\end{array}
Step 5:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{5}{4}}{2}\right)^2=\frac{25}{64} \text{on }\text{the }\text{left }\text{and }8\times \frac{25}{64}=\frac{25}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 6:
\begin{array}{l}
x^2+\frac{5 x}{4}+\frac{25}{64}=\left(x+\frac{5}{8}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 8 \fbox{$\left(x+\frac{5}{8}\right)^2$}=\frac{25}{8}-3 y \\
\end{array}
| khanacademy | amps |
Given the equation $-10 x^2+9 x-10 y^2+3 y+7=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-10 y^2+3 y-10 x^2+9 x+7=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }7 \text{from }\text{both }\text{sides}: \\
-10 y^2+3 y-10 x^2+9 x=-7 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-10 x^2+9 x+\underline{\text{ }}\right)+\left(-10 y^2+3 y+\underline{\text{ }}\right)=\underline{\text{ }}-7 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-10 x^2+9 x+\underline{\text{ }}\right)=-10 \left(x^2-\frac{9 x}{10}+\underline{\text{ }}\right): \\
\fbox{$-10 \left(x^2-\frac{9 x}{10}+\underline{\text{ }}\right)$}+\left(-10 y^2+3 y+\underline{\text{ }}\right)=\underline{\text{ }}-7 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-10 y^2+3 y+\underline{\text{ }}\right)=-10 \left(y^2-\frac{3 y}{10}+\underline{\text{ }}\right): \\
-10 \left(x^2-\frac{9 x}{10}+\underline{\text{ }}\right)+\fbox{$-10 \left(y^2-\frac{3 y}{10}+\underline{\text{ }}\right)$}=\underline{\text{ }}-7 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-10 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-9}{10}}{2}\right)^2=\frac{81}{400} \text{on }\text{the }\text{left }\text{and }-10\times \frac{81}{400}=-\frac{81}{40} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-7-\frac{81}{40}=-\frac{361}{40}: \\
-10 \left(x^2-\frac{9 x}{10}+\frac{81}{400}\right)-10 \left(y^2-\frac{3 y}{10}+\underline{\text{ }}\right)=\fbox{$-\frac{361}{40}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-10 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-3}{10}}{2}\right)^2=\frac{9}{400} \text{on }\text{the }\text{left }\text{and }-10\times \frac{9}{400}=-\frac{9}{40} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{361}{40}-\frac{9}{40}=-\frac{37}{4}: \\
-10 \left(x^2-\frac{9 x}{10}+\frac{81}{400}\right)-10 \left(y^2-\frac{3 y}{10}+\frac{9}{400}\right)=\fbox{$-\frac{37}{4}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{9 x}{10}+\frac{81}{400}=\left(x-\frac{9}{20}\right)^2: \\
-10 \fbox{$\left(x-\frac{9}{20}\right)^2$}-10 \left(y^2-\frac{3 y}{10}+\frac{9}{400}\right)=-\frac{37}{4} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{3 y}{10}+\frac{9}{400}=\left(y-\frac{3}{20}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -10 \left(x-\frac{9}{20}\right)^2-\text{10 }\fbox{$\left(y-\frac{3}{20}\right)^2$}=-\frac{37}{4} \\
\end{array}
| khanacademy | amps |
Given the equation $-5 x^2-3 x-2 y^2+y-8=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-2 y^2+y-5 x^2-3 x-8=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }8 \text{to }\text{both }\text{sides}: \\
-2 y^2+y-5 x^2-3 x=8 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-5 x^2-3 x+\underline{\text{ }}\right)+\left(-2 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}+8 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-5 x^2-3 x+\underline{\text{ }}\right)=-5 \left(x^2+\frac{3 x}{5}+\underline{\text{ }}\right): \\
\fbox{$-5 \left(x^2+\frac{3 x}{5}+\underline{\text{ }}\right)$}+\left(-2 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}+8 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-2 y^2+y+\underline{\text{ }}\right)=-2 \left(y^2-\frac{y}{2}+\underline{\text{ }}\right): \\
-5 \left(x^2+\frac{3 x}{5}+\underline{\text{ }}\right)+\fbox{$-2 \left(y^2-\frac{y}{2}+\underline{\text{ }}\right)$}=\underline{\text{ }}+8 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{3}{5}}{2}\right)^2=\frac{9}{100} \text{on }\text{the }\text{left }\text{and }-5\times \frac{9}{100}=-\frac{9}{20} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
8-\frac{9}{20}=\frac{151}{20}: \\
-5 \left(x^2+\frac{3 x}{5}+\frac{9}{100}\right)-2 \left(y^2-\frac{y}{2}+\underline{\text{ }}\right)=\fbox{$\frac{151}{20}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-1}{2}}{2}\right)^2=\frac{1}{16} \text{on }\text{the }\text{left }\text{and }\frac{-2}{16}=-\frac{1}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{151}{20}-\frac{1}{8}=\frac{297}{40}: \\
-5 \left(x^2+\frac{3 x}{5}+\frac{9}{100}\right)-2 \left(y^2-\frac{y}{2}+\frac{1}{16}\right)=\fbox{$\frac{297}{40}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{3 x}{5}+\frac{9}{100}=\left(x+\frac{3}{10}\right)^2: \\
-5 \fbox{$\left(x+\frac{3}{10}\right)^2$}-2 \left(y^2-\frac{y}{2}+\frac{1}{16}\right)=\frac{297}{40} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{y}{2}+\frac{1}{16}=\left(y-\frac{1}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -5 \left(x+\frac{3}{10}\right)^2-2 \fbox{$\left(y-\frac{1}{4}\right)^2$}=\frac{297}{40} \\
\end{array}
| khanacademy | amps |
Given the equation $2 x^2-x+7 y^2-6 y-10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
7 y^2-6 y+2 x^2-x-10=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }\text{10 }\text{to }\text{both }\text{sides}: \\
7 y^2-6 y+2 x^2-x=10 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(2 x^2-x+\underline{\text{ }}\right)+\left(7 y^2-6 y+\underline{\text{ }}\right)=\underline{\text{ }}+10 \\
\end{array}
Step 4:
\begin{array}{l}
\left(2 x^2-x+\underline{\text{ }}\right)=2 \left(x^2-\frac{x}{2}+\underline{\text{ }}\right): \\
\fbox{$2 \left(x^2-\frac{x}{2}+\underline{\text{ }}\right)$}+\left(7 y^2-6 y+\underline{\text{ }}\right)=\underline{\text{ }}+10 \\
\end{array}
Step 5:
\begin{array}{l}
\left(7 y^2-6 y+\underline{\text{ }}\right)=7 \left(y^2-\frac{6 y}{7}+\underline{\text{ }}\right): \\
2 \left(x^2-\frac{x}{2}+\underline{\text{ }}\right)+\fbox{$7 \left(y^2-\frac{6 y}{7}+\underline{\text{ }}\right)$}=\underline{\text{ }}+10 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-1}{2}}{2}\right)^2=\frac{1}{16} \text{on }\text{the }\text{left }\text{and }\frac{2}{16}=\frac{1}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
10+\frac{1}{8}=\frac{81}{8}: \\
2 \left(x^2-\frac{x}{2}+\frac{1}{16}\right)+7 \left(y^2-\frac{6 y}{7}+\underline{\text{ }}\right)=\fbox{$\frac{81}{8}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-6}{7}}{2}\right)^2=\frac{9}{49} \text{on }\text{the }\text{left }\text{and }7\times \frac{9}{49}=\frac{9}{7} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{81}{8}+\frac{9}{7}=\frac{639}{56}: \\
2 \left(x^2-\frac{x}{2}+\frac{1}{16}\right)+7 \left(y^2-\frac{6 y}{7}+\frac{9}{49}\right)=\fbox{$\frac{639}{56}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{x}{2}+\frac{1}{16}=\left(x-\frac{1}{4}\right)^2: \\
2 \fbox{$\left(x-\frac{1}{4}\right)^2$}+7 \left(y^2-\frac{6 y}{7}+\frac{9}{49}\right)=\frac{639}{56} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{6 y}{7}+\frac{9}{49}=\left(y-\frac{3}{7}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 2 \left(x-\frac{1}{4}\right)^2+7 \fbox{$\left(y-\frac{3}{7}\right)^2$}=\frac{639}{56} \\
\end{array}
| khanacademy | amps |
Given the equation $2 x^2-9 x+2 y^2+9 y-6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
2 y^2+9 y+2 x^2-9 x-6=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }6 \text{to }\text{both }\text{sides}: \\
2 y^2+9 y+2 x^2-9 x=6 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(2 x^2-9 x+\underline{\text{ }}\right)+\left(2 y^2+9 y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\end{array}
Step 4:
\begin{array}{l}
\left(2 x^2-9 x+\underline{\text{ }}\right)=2 \left(x^2-\frac{9 x}{2}+\underline{\text{ }}\right): \\
\fbox{$2 \left(x^2-\frac{9 x}{2}+\underline{\text{ }}\right)$}+\left(2 y^2+9 y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\end{array}
Step 5:
\begin{array}{l}
\left(2 y^2+9 y+\underline{\text{ }}\right)=2 \left(y^2+\frac{9 y}{2}+\underline{\text{ }}\right): \\
2 \left(x^2-\frac{9 x}{2}+\underline{\text{ }}\right)+\fbox{$2 \left(y^2+\frac{9 y}{2}+\underline{\text{ }}\right)$}=\underline{\text{ }}+6 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-9}{2}}{2}\right)^2=\frac{81}{16} \text{on }\text{the }\text{left }\text{and }2\times \frac{81}{16}=\frac{81}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
6+\frac{81}{8}=\frac{129}{8}: \\
2 \left(x^2-\frac{9 x}{2}+\frac{81}{16}\right)+2 \left(y^2+\frac{9 y}{2}+\underline{\text{ }}\right)=\fbox{$\frac{129}{8}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{9}{2}}{2}\right)^2=\frac{81}{16} \text{on }\text{the }\text{left }\text{and }2\times \frac{81}{16}=\frac{81}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{129}{8}+\frac{81}{8}=\frac{105}{4}: \\
2 \left(x^2-\frac{9 x}{2}+\frac{81}{16}\right)+2 \left(y^2+\frac{9 y}{2}+\frac{81}{16}\right)=\fbox{$\frac{105}{4}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{9 x}{2}+\frac{81}{16}=\left(x-\frac{9}{4}\right)^2: \\
2 \fbox{$\left(x-\frac{9}{4}\right)^2$}+2 \left(y^2+\frac{9 y}{2}+\frac{81}{16}\right)=\frac{105}{4} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{9 y}{2}+\frac{81}{16}=\left(y+\frac{9}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 2 \left(x-\frac{9}{4}\right)^2+2 \fbox{$\left(y+\frac{9}{4}\right)^2$}=\frac{105}{4} \\
\end{array}
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