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Given the equation $-5 x^2-7 x-4 y^2+10 y+10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-4 y^2+10 y-5 x^2-7 x+10=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }\text{10 }\text{from }\text{both }\text{sides}: \\
-4 y^2+10 y-5 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(-5 x^2-7 x+\underline{\text{ }}\right)+\left(-4 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}-10 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-5 x^2-7 x+\underline{\text{ }}\right)=-5 \left(x^2+\frac{7 x}{5}+\underline{\text{ }}\right): \\
\fbox{$-5 \left(x^2+\frac{7 x}{5}+\underline{\text{ }}\right)$}+\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): \\
-5 \left(x^2+\frac{7 x}{5}+\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 }-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-\frac{49}{20}=-\frac{249}{20}: \\
-5 \left(x^2+\frac{7 x}{5}+\frac{49}{100}\right)-4 \left(y^2-\frac{5 y}{2}+\underline{\text{ }}\right)=\fbox{$-\frac{249}{20}$} \\
\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 }-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{249}{20}-\frac{25}{4}=-\frac{187}{10}: \\
-5 \left(x^2+\frac{7 x}{5}+\frac{49}{100}\right)-4 \left(y^2-\frac{5 y}{2}+\frac{25}{16}\right)=\fbox{$-\frac{187}{10}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{7 x}{5}+\frac{49}{100}=\left(x+\frac{7}{10}\right)^2: \\
-5 \fbox{$\left(x+\frac{7}{10}\right)^2$}-4 \left(y^2-\frac{5 y}{2}+\frac{25}{16}\right)=-\frac{187}{10} \\
\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{} & -5 \left(x+\frac{7}{10}\right)^2-4 \fbox{$\left(y-\frac{5}{4}\right)^2$}=-\frac{187}{10} \\
\end{array}
| khanacademy | amps |
Given the equation $8 x^2-10 x+9 y+3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
8 x^2-10 x+(9 y+3)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }9 y+3 \text{from }\text{both }\text{sides}: \\
8 x^2-10 x=-9 y-3 \\
\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)=(-9 y-3)+\underline{\text{ }} \\
\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{$8 \left(x^2-\frac{5 x}{4}+\underline{\text{ }}\right)$}=(-9 y-3)+\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 }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}
(-9 y-3)+\frac{25}{8}=\frac{1}{8}-9 y: \\
8 \left(x^2-\frac{5 x}{4}+\frac{25}{64}\right)=\fbox{$\frac{1}{8}-9 y$} \\
\end{array}
Step 7:
\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{1}{8}-9 y \\
\end{array}
| khanacademy | amps |
Given the equation $-7 x^2-6 x-2 y^2+10 y+6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-2 y^2+10 y-7 x^2-6 x+6=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }6 \text{from }\text{both }\text{sides}: \\
-2 y^2+10 y-7 x^2-6 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^2-6 x+\underline{\text{ }}\right)+\left(-2 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}-6 \\
\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(-2 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}-6 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-2 y^2+10 y+\underline{\text{ }}\right)=-2 \left(y^2-5 y+\underline{\text{ }}\right): \\
-7 \left(x^2+\frac{6 x}{7}+\underline{\text{ }}\right)+\fbox{$-2 \left(y^2-5 y+\underline{\text{ }}\right)$}=\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 }-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}
-6-\frac{9}{7}=-\frac{51}{7}: \\
-7 \left(x^2+\frac{6 x}{7}+\frac{9}{49}\right)-2 \left(y^2-5 y+\underline{\text{ }}\right)=\fbox{$-\frac{51}{7}$} \\
\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{-5}{2}\right)^2=\frac{25}{4} \text{on }\text{the }\text{left }\text{and }-2\times \frac{25}{4}=-\frac{25}{2} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{51}{7}-\frac{25}{2}=-\frac{277}{14}: \\
-7 \left(x^2+\frac{6 x}{7}+\frac{9}{49}\right)-2 \left(y^2-5 y+\frac{25}{4}\right)=\fbox{$-\frac{277}{14}$} \\
\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{$\left(x+\frac{3}{7}\right)^2$}-2 \left(y^2-5 y+\frac{25}{4}\right)=-\frac{277}{14} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-5 y+\frac{25}{4}=\left(y-\frac{5}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -7 \left(x+\frac{3}{7}\right)^2-2 \fbox{$\left(y-\frac{5}{2}\right)^2$}=-\frac{277}{14} \\
\end{array}
| khanacademy | amps |
Given the equation $6 x^2+6 x+6 y^2-8 y=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
6 y^2-8 y+6 x^2+6 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(6 x^2+6 x+\underline{\text{ }}\right)+\left(6 y^2-8 y+\underline{\text{ }}\right)=\underline{\text{ }}+0 \\
\end{array}
Step 3:
\begin{array}{l}
\left(6 x^2+6 x+\underline{\text{ }}\right)=6 \left(x^2+x+\underline{\text{ }}\right): \\
\fbox{$6 \left(x^2+x+\underline{\text{ }}\right)$}+\left(6 y^2-8 y+\underline{\text{ }}\right)=\underline{\text{ }}+0 \\
\end{array}
Step 4:
\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+x+\underline{\text{ }}\right)+\fbox{$6 \left(y^2-\frac{4 y}{3}+\underline{\text{ }}\right)$}=\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 }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 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 }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}
\frac{3}{2}+\frac{8}{3}=\frac{25}{6}: \\
6 \left(x^2+x+\frac{1}{4}\right)+6 \left(y^2-\frac{4 y}{3}+\frac{4}{9}\right)=\fbox{$\frac{25}{6}$} \\
\end{array}
Step 8:
\begin{array}{l}
x^2+x+\frac{1}{4}=\left(x+\frac{1}{2}\right)^2: \\
6 \fbox{$\left(x+\frac{1}{2}\right)^2$}+6 \left(y^2-\frac{4 y}{3}+\frac{4}{9}\right)=\frac{25}{6} \\
\end{array}
Step 9:
\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}{2}\right)^2+6 \fbox{$\left(y-\frac{2}{3}\right)^2$}=\frac{25}{6} \\
\end{array}
| khanacademy | amps |
Given the equation $-3 x^2+x-6 y^2-9 y-8=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-6 y^2-9 y-3 x^2+x-8=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }8 \text{to }\text{both }\text{sides}: \\
-6 y^2-9 y-3 x^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(-3 x^2+x+\underline{\text{ }}\right)+\left(-6 y^2-9 y+\underline{\text{ }}\right)=\underline{\text{ }}+8 \\
\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(-6 y^2-9 y+\underline{\text{ }}\right)=\underline{\text{ }}+8 \\
\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): \\
-3 \left(x^2-\frac{x}{3}+\underline{\text{ }}\right)+\fbox{$-6 \left(y^2+\frac{3 y}{2}+\underline{\text{ }}\right)$}=\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 }-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}
8-\frac{1}{12}=\frac{95}{12}: \\
-3 \left(x^2-\frac{x}{3}+\frac{1}{36}\right)-6 \left(y^2+\frac{3 y}{2}+\underline{\text{ }}\right)=\fbox{$\frac{95}{12}$} \\
\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 }-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{95}{12}-\frac{27}{8}=\frac{109}{24}: \\
-3 \left(x^2-\frac{x}{3}+\frac{1}{36}\right)-6 \left(y^2+\frac{3 y}{2}+\frac{9}{16}\right)=\fbox{$\frac{109}{24}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{x}{3}+\frac{1}{36}=\left(x-\frac{1}{6}\right)^2: \\
-3 \fbox{$\left(x-\frac{1}{6}\right)^2$}-6 \left(y^2+\frac{3 y}{2}+\frac{9}{16}\right)=\frac{109}{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{1}{6}\right)^2-6 \fbox{$\left(y+\frac{3}{4}\right)^2$}=\frac{109}{24} \\
\end{array}
| khanacademy | amps |
Given the equation $9 x^2-8 x-9 y^2-6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
9 x^2-8 x+\left(-9 y^2-6\right)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }6 \text{to }\text{both }\text{sides}: \\
-9 y^2+9 x^2-8 x=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^2-8 x+\underline{\text{ }}\right)-9 y^2=\underline{\text{ }}+6 \\
\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{$9 \left(x^2-\frac{8 x}{9}+\underline{\text{ }}\right)$}-9 y^2=\underline{\text{ }}+6 \\
\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{-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 6:
\begin{array}{l}
6+\frac{16}{9}=\frac{70}{9}: \\
9 \left(x^2-\frac{8 x}{9}+\frac{16}{81}\right)-9 y^2=\fbox{$\frac{70}{9}$} \\
\end{array}
Step 7:
\begin{array}{l}
x^2-\frac{8 x}{9}+\frac{16}{81}=\left(x-\frac{4}{9}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 9 \fbox{$\left(x-\frac{4}{9}\right)^2$}-9 y^2=\frac{70}{9} \\
\end{array}
| khanacademy | amps |
Given the equation $x^2-2 y^2-2 y=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-2 y^2-2 y+x^2=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }-2 y^2-2 y+x^2 \text{from }\text{both }\text{sides}: \\
2 y^2+2 y-x^2=0 \\
\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(2 y^2+2 y+\underline{\text{ }}\right)-x^2=\underline{\text{ }}+0 \\
\end{array}
Step 4:
\begin{array}{l}
\left(2 y^2+2 y+\underline{\text{ }}\right)=2 \left(y^2+y+\underline{\text{ }}\right): \\
\fbox{$2 \left(y^2+y+\underline{\text{ }}\right)$}-x^2=\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 }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 6:
\begin{array}{l}
y^2+y+\frac{1}{4}=\left(y+\frac{1}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 2 \fbox{$\left(y+\frac{1}{2}\right)^2$}-x^2=\frac{1}{2} \\
\end{array}
| khanacademy | amps |
Given the equation $7 x^2+4 x+6 y^2+6 y+8=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
6 y^2+6 y+7 x^2+4 x+8=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }8 \text{from }\text{both }\text{sides}: \\
6 y^2+6 y+7 x^2+4 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(7 x^2+4 x+\underline{\text{ }}\right)+\left(6 y^2+6 y+\underline{\text{ }}\right)=\underline{\text{ }}-8 \\
\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(6 y^2+6 y+\underline{\text{ }}\right)=\underline{\text{ }}-8 \\
\end{array}
Step 5:
\begin{array}{l}
\left(6 y^2+6 y+\underline{\text{ }}\right)=6 \left(y^2+y+\underline{\text{ }}\right): \\
7 \left(x^2+\frac{4 x}{7}+\underline{\text{ }}\right)+\fbox{$6 \left(y^2+y+\underline{\text{ }}\right)$}=\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 }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}
\frac{4}{7}-8=-\frac{52}{7}: \\
7 \left(x^2+\frac{4 x}{7}+\frac{4}{49}\right)+6 \left(y^2+y+\underline{\text{ }}\right)=\fbox{$-\frac{52}{7}$} \\
\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{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{3}{2}-\frac{52}{7}=-\frac{83}{14}: \\
7 \left(x^2+\frac{4 x}{7}+\frac{4}{49}\right)+6 \left(y^2+y+\frac{1}{4}\right)=\fbox{$-\frac{83}{14}$} \\
\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{$\left(x+\frac{2}{7}\right)^2$}+6 \left(y^2+y+\frac{1}{4}\right)=-\frac{83}{14} \\
\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{} & 7 \left(x+\frac{2}{7}\right)^2+6 \fbox{$\left(y+\frac{1}{2}\right)^2$}=-\frac{83}{14} \\
\end{array}
| khanacademy | amps |
Given the equation $7 x^2+6 x-5 y^2+3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
7 x^2+6 x+\left(3-5 y^2\right)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }3 \text{from }\text{both }\text{sides}: \\
-5 y^2+7 x^2+6 x=-3 \\
\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 x^2+6 x+\underline{\text{ }}\right)-5 y^2=\underline{\text{ }}-3 \\
\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{$7 \left(x^2+\frac{6 x}{7}+\underline{\text{ }}\right)$}-5 y^2=\underline{\text{ }}-3 \\
\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{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 6:
\begin{array}{l}
\frac{9}{7}-3=-\frac{12}{7}: \\
7 \left(x^2+\frac{6 x}{7}+\frac{9}{49}\right)-5 y^2=\fbox{$-\frac{12}{7}$} \\
\end{array}
Step 7:
\begin{array}{l}
x^2+\frac{6 x}{7}+\frac{9}{49}=\left(x+\frac{3}{7}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 7 \fbox{$\left(x+\frac{3}{7}\right)^2$}-5 y^2=-\frac{12}{7} \\
\end{array}
| khanacademy | amps |
Given the equation $-3 x^2-10 x+3 y^2+8=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-3 x^2-10 x+\left(3 y^2+8\right)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }3 y^2-3 x^2-10 x+8 \text{from }\text{both }\text{sides}: \\
3 x^2+10 x+\left(-3 y^2-8\right)=0 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Add }8 \text{to }\text{both }\text{sides}: \\
-3 y^2+3 x^2+10 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(3 x^2+10 x+\underline{\text{ }}\right)-3 y^2=\underline{\text{ }}+8 \\
\end{array}
Step 5:
\begin{array}{l}
\left(3 x^2+10 x+\underline{\text{ }}\right)=3 \left(x^2+\frac{10 x}{3}+\underline{\text{ }}\right): \\
\fbox{$3 \left(x^2+\frac{10 x}{3}+\underline{\text{ }}\right)$}-3 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 }3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{10}{3}}{2}\right)^2=\frac{25}{9} \text{on }\text{the }\text{left }\text{and }3\times \frac{25}{9}=\frac{25}{3} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
8+\frac{25}{3}=\frac{49}{3}: \\
3 \left(x^2+\frac{10 x}{3}+\frac{25}{9}\right)-3 y^2=\fbox{$\frac{49}{3}$} \\
\end{array}
Step 8:
\begin{array}{l}
x^2+\frac{10 x}{3}+\frac{25}{9}=\left(x+\frac{5}{3}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 3 \fbox{$\left(x+\frac{5}{3}\right)^2$}-3 y^2=\frac{49}{3} \\
\end{array}
| khanacademy | amps |
Given the equation $9 x^2+x-8 y^2+9 y+4=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-8 y^2+9 y+9 x^2+x+4=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }4 \text{from }\text{both }\text{sides}: \\
-8 y^2+9 y+9 x^2+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+x+\underline{\text{ }}\right)+\left(-8 y^2+9 y+\underline{\text{ }}\right)=\underline{\text{ }}-4 \\
\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{$9 \left(x^2+\frac{x}{9}+\underline{\text{ }}\right)$}+\left(-8 y^2+9 y+\underline{\text{ }}\right)=\underline{\text{ }}-4 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-8 y^2+9 y+\underline{\text{ }}\right)=-8 \left(y^2-\frac{9 y}{8}+\underline{\text{ }}\right): \\
9 \left(x^2+\frac{x}{9}+\underline{\text{ }}\right)+\fbox{$-8 \left(y^2-\frac{9 y}{8}+\underline{\text{ }}\right)$}=\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{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 7:
\begin{array}{l}
\frac{1}{36}-4=-\frac{143}{36}: \\
9 \left(x^2+\frac{x}{9}+\frac{1}{324}\right)-8 \left(y^2-\frac{9 y}{8}+\underline{\text{ }}\right)=\fbox{$-\frac{143}{36}$} \\
\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{-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 9:
\begin{array}{l}
-\frac{143}{36}-\frac{81}{32}=-\frac{1873}{288}: \\
9 \left(x^2+\frac{x}{9}+\frac{1}{324}\right)-8 \left(y^2-\frac{9 y}{8}+\frac{81}{256}\right)=\fbox{$-\frac{1873}{288}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{x}{9}+\frac{1}{324}=\left(x+\frac{1}{18}\right)^2: \\
9 \fbox{$\left(x+\frac{1}{18}\right)^2$}-8 \left(y^2-\frac{9 y}{8}+\frac{81}{256}\right)=-\frac{1873}{288} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{9 y}{8}+\frac{81}{256}=\left(y-\frac{9}{16}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 9 \left(x+\frac{1}{18}\right)^2-8 \fbox{$\left(y-\frac{9}{16}\right)^2$}=-\frac{1873}{288} \\
\end{array}
| khanacademy | amps |
Given the equation $-4 x^2+8 x-6 y^2+8 y-5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-6 y^2+8 y-4 x^2+8 x-5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }5 \text{to }\text{both }\text{sides}: \\
-6 y^2+8 y-4 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(-4 x^2+8 x+\underline{\text{ }}\right)+\left(-6 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}+5 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-4 x^2+8 x+\underline{\text{ }}\right)=-4 \left(x^2-2 x+\underline{\text{ }}\right): \\
\fbox{$-4 \left(x^2-2 x+\underline{\text{ }}\right)$}+\left(-6 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}+5 \\
\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): \\
-4 \left(x^2-2 x+\underline{\text{ }}\right)+\fbox{$-6 \left(y^2-\frac{4 y}{3}+\underline{\text{ }}\right)$}=\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 }-4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-2}{2}\right)^2=1 \text{on }\text{the }\text{left }\text{and }-4\times 1=-4 \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
5-4=1: \\
-4 \left(x^2-2 x+1\right)-6 \left(y^2-\frac{4 y}{3}+\underline{\text{ }}\right)=\fbox{$1$} \\
\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}
1-\frac{8}{3}=-\frac{5}{3}: \\
-4 \left(x^2-2 x+1\right)-6 \left(y^2-\frac{4 y}{3}+\frac{4}{9}\right)=\fbox{$-\frac{5}{3}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-2 x+1=(x-1)^2: \\
-4 \fbox{$(x-1)^2$}-6 \left(y^2-\frac{4 y}{3}+\frac{4}{9}\right)=-\frac{5}{3} \\
\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{} & -4 (x-1)^2-6 \fbox{$\left(y-\frac{2}{3}\right)^2$}=-\frac{5}{3} \\
\end{array}
| khanacademy | amps |
Given the equation $-x^2+2 x-4 y^2+8=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-x^2+2 x+\left(8-4 y^2\right)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }-4 y^2-x^2+2 x+8 \text{from }\text{both }\text{sides}: \\
x^2-2 x+\left(4 y^2-8\right)=0 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Add }8 \text{to }\text{both }\text{sides}: \\
4 y^2+x^2-2 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(x^2-2 x+\underline{\text{ }}\right)+4 y^2=\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{Add }\left(\frac{-2}{2}\right)^2=1 \text{to }\text{both }\text{sides}: \\
\end{array}
Step 6:
\begin{array}{l}
8+1=9: \\
\left(x^2-2 x+1\right)+4 y^2=\fbox{$9$} \\
\end{array}
Step 7:
\begin{array}{l}
x^2-2 x+1=(x-1)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & \fbox{$(x-1)^2$}+4 y^2=9 \\
\end{array}
| khanacademy | amps |
Given the equation $-8 x^2+6 x-y^2+5 y-4=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-y^2+5 y-8 x^2+6 x-4=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }4 \text{to }\text{both }\text{sides}: \\
-y^2+5 y-8 x^2+6 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(-8 x^2+6 x+\underline{\text{ }}\right)+\left(-y^2+5 y+\underline{\text{ }}\right)=\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{$-8 \left(x^2-\frac{3 x}{4}+\underline{\text{ }}\right)$}+\left(-y^2+5 y+\underline{\text{ }}\right)=\underline{\text{ }}+4 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-y^2+5 y+\underline{\text{ }}\right)=-\left(y^2-5 y+\underline{\text{ }}\right): \\
-8 \left(x^2-\frac{3 x}{4}+\underline{\text{ }}\right)+\fbox{$-\left(y^2-5 y+\underline{\text{ }}\right)$}=\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 }-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}
4-\frac{9}{8}=\frac{23}{8}: \\
-8 \left(x^2-\frac{3 x}{4}+\frac{9}{64}\right)-\left(y^2-5 y+\underline{\text{ }}\right)=\fbox{$\frac{23}{8}$} \\
\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 }-1 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-5}{2}\right)^2=\frac{25}{4} \text{on }\text{the }\text{left }\text{and }-\frac{25}{4}=-\frac{25}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{23}{8}-\frac{25}{4}=-\frac{27}{8}: \\
-8 \left(x^2-\frac{3 x}{4}+\frac{9}{64}\right)-\left(y^2-5 y+\frac{25}{4}\right)=\fbox{$-\frac{27}{8}$} \\
\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{$\left(x-\frac{3}{8}\right)^2$}-\left(y^2-5 y+\frac{25}{4}\right)=-\frac{27}{8} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-5 y+\frac{25}{4}=\left(y-\frac{5}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -8 \left(x-\frac{3}{8}\right)^2-\fbox{$\left(y-\frac{5}{2}\right)^2$}=-\frac{27}{8} \\
\end{array}
| khanacademy | amps |
Given the equation $-2 x^2-2 x+7 y^2+6 y+1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
7 y^2+6 y-2 x^2-2 x+1=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }1 \text{from }\text{both }\text{sides}: \\
7 y^2+6 y-2 x^2-2 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(-2 x^2-2 x+\underline{\text{ }}\right)+\left(7 y^2+6 y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\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+6 y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\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+x+\underline{\text{ }}\right)+\fbox{$7 \left(y^2+\frac{6 y}{7}+\underline{\text{ }}\right)$}=\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 }-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}
-1-\frac{1}{2}=-\frac{3}{2}: \\
-2 \left(x^2+x+\frac{1}{4}\right)+7 \left(y^2+\frac{6 y}{7}+\underline{\text{ }}\right)=\fbox{$-\frac{3}{2}$} \\
\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{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{3}{2}=-\frac{3}{14}: \\
-2 \left(x^2+x+\frac{1}{4}\right)+7 \left(y^2+\frac{6 y}{7}+\frac{9}{49}\right)=\fbox{$-\frac{3}{14}$} \\
\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{6 y}{7}+\frac{9}{49}\right)=-\frac{3}{14} \\
\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}{2}\right)^2+7 \fbox{$\left(y+\frac{3}{7}\right)^2$}=-\frac{3}{14} \\
\end{array}
| khanacademy | amps |
Given the equation $8 x^2+6 x-9 y^2+6 y+7=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-9 y^2+6 y+8 x^2+6 x+7=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }7 \text{from }\text{both }\text{sides}: \\
-9 y^2+6 y+8 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(8 x^2+6 x+\underline{\text{ }}\right)+\left(-9 y^2+6 y+\underline{\text{ }}\right)=\underline{\text{ }}-7 \\
\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(-9 y^2+6 y+\underline{\text{ }}\right)=\underline{\text{ }}-7 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-9 y^2+6 y+\underline{\text{ }}\right)=-9 \left(y^2-\frac{2 y}{3}+\underline{\text{ }}\right): \\
8 \left(x^2+\frac{3 x}{4}+\underline{\text{ }}\right)+\fbox{$-9 \left(y^2-\frac{2 y}{3}+\underline{\text{ }}\right)$}=\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 }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}
\frac{9}{8}-7=-\frac{47}{8}: \\
8 \left(x^2+\frac{3 x}{4}+\frac{9}{64}\right)-9 \left(y^2-\frac{2 y}{3}+\underline{\text{ }}\right)=\fbox{$-\frac{47}{8}$} \\
\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}{3}}{2}\right)^2=\frac{1}{9} \text{on }\text{the }\text{left }\text{and }\frac{-9}{9}=-1 \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{47}{8}-1=-\frac{55}{8}: \\
8 \left(x^2+\frac{3 x}{4}+\frac{9}{64}\right)-9 \left(y^2-\frac{2 y}{3}+\frac{1}{9}\right)=\fbox{$-\frac{55}{8}$} \\
\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{$\left(x+\frac{3}{8}\right)^2$}-9 \left(y^2-\frac{2 y}{3}+\frac{1}{9}\right)=-\frac{55}{8} \\
\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{} & 8 \left(x+\frac{3}{8}\right)^2-9 \fbox{$\left(y-\frac{1}{3}\right)^2$}=-\frac{55}{8} \\
\end{array}
| khanacademy | amps |
Given the equation $x^2+5 x+9 y^2-6 y+5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
9 y^2-6 y+x^2+5 x+5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }5 \text{from }\text{both }\text{sides}: \\
9 y^2-6 y+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(x^2+5 x+\underline{\text{ }}\right)+\left(9 y^2-6 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\end{array}
Step 4:
\begin{array}{l}
\left(9 y^2-6 y+\underline{\text{ }}\right)=9 \left(y^2-\frac{2 y}{3}+\underline{\text{ }}\right): \\
\left(x^2+5 x+\underline{\text{ }}\right)+\fbox{$9 \left(y^2-\frac{2 y}{3}+\underline{\text{ }}\right)$}=\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{5}{2}\right)^2=\frac{25}{4} \text{to }\text{both }\text{sides}: \\
\end{array}
Step 6:
\begin{array}{l}
\frac{25}{4}-5=\frac{5}{4}: \\
\left(x^2+5 x+\frac{25}{4}\right)+9 \left(y^2-\frac{2 y}{3}+\underline{\text{ }}\right)=\fbox{$\frac{5}{4}$} \\
\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 }9 \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{9}{9}=1 \text{on }\text{the }\text{right}: \\
\end{array}
Step 8:
\begin{array}{l}
\frac{5}{4}+1=\frac{9}{4}: \\
\left(x^2+5 x+\frac{25}{4}\right)+9 \left(y^2-\frac{2 y}{3}+\frac{1}{9}\right)=\fbox{$\frac{9}{4}$} \\
\end{array}
Step 9:
\begin{array}{l}
x^2+5 x+\frac{25}{4}=\left(x+\frac{5}{2}\right)^2: \\
\fbox{$\left(x+\frac{5}{2}\right)^2$}+9 \left(y^2-\frac{2 y}{3}+\frac{1}{9}\right)=\frac{9}{4} \\
\end{array}
Step 10:
\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{5}{2}\right)^2+9 \fbox{$\left(y-\frac{1}{3}\right)^2$}=\frac{9}{4} \\
\end{array}
| khanacademy | amps |
Given the equation $-x^2+4 x+10 y^2-3 y+4=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
10 y^2-3 y-x^2+4 x+4=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }4 \text{from }\text{both }\text{sides}: \\
10 y^2-3 y-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(-x^2+4 x+\underline{\text{ }}\right)+\left(10 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}-4 \\
\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(10 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}-4 \\
\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): \\
-\left(x^2-4 x+\underline{\text{ }}\right)+\fbox{$10 \left(y^2-\frac{3 y}{10}+\underline{\text{ }}\right)$}=\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 }-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}
-4-4=-8: \\
-\left(x^2-4 x+4\right)+10 \left(y^2-\frac{3 y}{10}+\underline{\text{ }}\right)=\fbox{$-8$} \\
\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{-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{9}{40}-8=-\frac{311}{40}: \\
-\left(x^2-4 x+4\right)+10 \left(y^2-\frac{3 y}{10}+\frac{9}{400}\right)=\fbox{$-\frac{311}{40}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-4 x+4=(x-2)^2: \\
-\fbox{$(x-2)^2$}+10 \left(y^2-\frac{3 y}{10}+\frac{9}{400}\right)=-\frac{311}{40} \\
\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{} & -(x-2)^2+\text{10 }\fbox{$\left(y-\frac{3}{20}\right)^2$}=-\frac{311}{40} \\
\end{array}
| khanacademy | amps |
Given the equation $-5 x^2+10 x+2 y^2+5 y+10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
2 y^2+5 y-5 x^2+10 x+10=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }\text{10 }\text{from }\text{both }\text{sides}: \\
2 y^2+5 y-5 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(-5 x^2+10 x+\underline{\text{ }}\right)+\left(2 y^2+5 y+\underline{\text{ }}\right)=\underline{\text{ }}-10 \\
\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(2 y^2+5 y+\underline{\text{ }}\right)=\underline{\text{ }}-10 \\
\end{array}
Step 5:
\begin{array}{l}
\left(2 y^2+5 y+\underline{\text{ }}\right)=2 \left(y^2+\frac{5 y}{2}+\underline{\text{ }}\right): \\
-5 \left(x^2-2 x+\underline{\text{ }}\right)+\fbox{$2 \left(y^2+\frac{5 y}{2}+\underline{\text{ }}\right)$}=\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 }-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}
-10-5=-15: \\
-5 \left(x^2-2 x+1\right)+2 \left(y^2+\frac{5 y}{2}+\underline{\text{ }}\right)=\fbox{$-15$} \\
\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{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 9:
\begin{array}{l}
\frac{25}{8}-15=-\frac{95}{8}: \\
-5 \left(x^2-2 x+1\right)+2 \left(y^2+\frac{5 y}{2}+\frac{25}{16}\right)=\fbox{$-\frac{95}{8}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-2 x+1=(x-1)^2: \\
-5 \fbox{$(x-1)^2$}+2 \left(y^2+\frac{5 y}{2}+\frac{25}{16}\right)=-\frac{95}{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{} & -5 (x-1)^2+2 \fbox{$\left(y+\frac{5}{4}\right)^2$}=-\frac{95}{8} \\
\end{array}
| khanacademy | amps |
Given the equation $9 x^2-3 x-7 y+10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
9 x^2-3 x+(10-7 y)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }10-7 y \text{from }\text{both }\text{sides}: \\
9 x^2-3 x=7 y-10 \\
\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^2-3 x+\underline{\text{ }}\right)=(7 y-10)+\underline{\text{ }} \\
\end{array}
Step 4:
\begin{array}{l}
\left(9 x^2-3 x+\underline{\text{ }}\right)=9 \left(x^2-\frac{x}{3}+\underline{\text{ }}\right): \\
\fbox{$9 \left(x^2-\frac{x}{3}+\underline{\text{ }}\right)$}=(7 y-10)+\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}{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 6:
\begin{array}{l}
(7 y-10)+\frac{1}{4}=7 y-\frac{39}{4}: \\
9 \left(x^2-\frac{x}{3}+\frac{1}{36}\right)=\fbox{$7 y-\frac{39}{4}$} \\
\end{array}
Step 7:
\begin{array}{l}
x^2-\frac{x}{3}+\frac{1}{36}=\left(x-\frac{1}{6}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 9 \fbox{$\left(x-\frac{1}{6}\right)^2$}=7 y-\frac{39}{4} \\
\end{array}
| khanacademy | amps |
Given the equation $6 x^2+5 x-4 y^2+2 y+5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-4 y^2+2 y+6 x^2+5 x+5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }5 \text{from }\text{both }\text{sides}: \\
-4 y^2+2 y+6 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(6 x^2+5 x+\underline{\text{ }}\right)+\left(-4 y^2+2 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\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(-4 y^2+2 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\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): \\
6 \left(x^2+\frac{5 x}{6}+\underline{\text{ }}\right)+\fbox{$-4 \left(y^2-\frac{y}{2}+\underline{\text{ }}\right)$}=\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{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}
\frac{25}{24}-5=-\frac{95}{24}: \\
6 \left(x^2+\frac{5 x}{6}+\frac{25}{144}\right)-4 \left(y^2-\frac{y}{2}+\underline{\text{ }}\right)=\fbox{$-\frac{95}{24}$} \\
\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 }-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{95}{24}-\frac{1}{4}=-\frac{101}{24}: \\
6 \left(x^2+\frac{5 x}{6}+\frac{25}{144}\right)-4 \left(y^2-\frac{y}{2}+\frac{1}{16}\right)=\fbox{$-\frac{101}{24}$} \\
\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{$\left(x+\frac{5}{12}\right)^2$}-4 \left(y^2-\frac{y}{2}+\frac{1}{16}\right)=-\frac{101}{24} \\
\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{} & 6 \left(x+\frac{5}{12}\right)^2-4 \fbox{$\left(y-\frac{1}{4}\right)^2$}=-\frac{101}{24} \\
\end{array}
| khanacademy | amps |
Given the equation $-5 x^2+5 y^2+5 y+7=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
5 y^2+5 y+\left(7-5 x^2\right)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }7 \text{from }\text{both }\text{sides}: \\
5 y^2+5 y-5 x^2=-7 \\
\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(5 y^2+5 y+\underline{\text{ }}\right)-5 x^2=\underline{\text{ }}-7 \\
\end{array}
Step 4:
\begin{array}{l}
\left(5 y^2+5 y+\underline{\text{ }}\right)=5 \left(y^2+y+\underline{\text{ }}\right): \\
\fbox{$5 \left(y^2+y+\underline{\text{ }}\right)$}-5 x^2=\underline{\text{ }}-7 \\
\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 }5 \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{5}{4}=\frac{5}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 6:
\begin{array}{l}
\frac{5}{4}-7=-\frac{23}{4}: \\
5 \left(y^2+y+\frac{1}{4}\right)-5 x^2=\fbox{$-\frac{23}{4}$} \\
\end{array}
Step 7:
\begin{array}{l}
y^2+y+\frac{1}{4}=\left(y+\frac{1}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 5 \fbox{$\left(y+\frac{1}{2}\right)^2$}-5 x^2=-\frac{23}{4} \\
\end{array}
| khanacademy | amps |
Given the equation $10 x^2-5 x+2 y^2+5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
10 x^2-5 x+\left(2 y^2+5\right)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }5 \text{from }\text{both }\text{sides}: \\
2 y^2+10 x^2-5 x=-5 \\
\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^2-5 x+\underline{\text{ }}\right)+2 y^2=\underline{\text{ }}-5 \\
\end{array}
Step 4:
\begin{array}{l}
\left(10 x^2-5 x+\underline{\text{ }}\right)=10 \left(x^2-\frac{x}{2}+\underline{\text{ }}\right): \\
\fbox{$10 \left(x^2-\frac{x}{2}+\underline{\text{ }}\right)$}+2 y^2=\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 }\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 6:
\begin{array}{l}
\frac{5}{8}-5=-\frac{35}{8}: \\
10 \left(x^2-\frac{x}{2}+\frac{1}{16}\right)+2 y^2=\fbox{$-\frac{35}{8}$} \\
\end{array}
Step 7:
\begin{array}{l}
x^2-\frac{x}{2}+\frac{1}{16}=\left(x-\frac{1}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & \text{10 }\fbox{$\left(x-\frac{1}{4}\right)^2$}+2 y^2=-\frac{35}{8} \\
\end{array}
| khanacademy | amps |
Given the equation $-2 x^2-6 x+7 y^2-7 y+3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
7 y^2-7 y-2 x^2-6 x+3=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }3 \text{from }\text{both }\text{sides}: \\
7 y^2-7 y-2 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(-2 x^2-6 x+\underline{\text{ }}\right)+\left(7 y^2-7 y+\underline{\text{ }}\right)=\underline{\text{ }}-3 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-2 x^2-6 x+\underline{\text{ }}\right)=-2 \left(x^2+3 x+\underline{\text{ }}\right): \\
\fbox{$-2 \left(x^2+3 x+\underline{\text{ }}\right)$}+\left(7 y^2-7 y+\underline{\text{ }}\right)=\underline{\text{ }}-3 \\
\end{array}
Step 5:
\begin{array}{l}
\left(7 y^2-7 y+\underline{\text{ }}\right)=7 \left(y^2-y+\underline{\text{ }}\right): \\
-2 \left(x^2+3 x+\underline{\text{ }}\right)+\fbox{$7 \left(y^2-y+\underline{\text{ }}\right)$}=\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 }-2 \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 }-2\times \frac{9}{4}=-\frac{9}{2} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-3-\frac{9}{2}=-\frac{15}{2}: \\
-2 \left(x^2+3 x+\frac{9}{4}\right)+7 \left(y^2-y+\underline{\text{ }}\right)=\fbox{$-\frac{15}{2}$} \\
\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{-1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{7}{4}=\frac{7}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{7}{4}-\frac{15}{2}=-\frac{23}{4}: \\
-2 \left(x^2+3 x+\frac{9}{4}\right)+7 \left(y^2-y+\frac{1}{4}\right)=\fbox{$-\frac{23}{4}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+3 x+\frac{9}{4}=\left(x+\frac{3}{2}\right)^2: \\
-2 \fbox{$\left(x+\frac{3}{2}\right)^2$}+7 \left(y^2-y+\frac{1}{4}\right)=-\frac{23}{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{} & -2 \left(x+\frac{3}{2}\right)^2+7 \fbox{$\left(y-\frac{1}{2}\right)^2$}=-\frac{23}{4} \\
\end{array}
| khanacademy | amps |
Given the equation $-6 x^2-2 x+2 y^2-3 y=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
2 y^2-3 y-6 x^2-2 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(-6 x^2-2 x+\underline{\text{ }}\right)+\left(2 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}+0 \\
\end{array}
Step 3:
\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(2 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}+0 \\
\end{array}
Step 4:
\begin{array}{l}
\left(2 y^2-3 y+\underline{\text{ }}\right)=2 \left(y^2-\frac{3 y}{2}+\underline{\text{ }}\right): \\
-6 \left(x^2+\frac{x}{3}+\underline{\text{ }}\right)+\fbox{$2 \left(y^2-\frac{3 y}{2}+\underline{\text{ }}\right)$}=\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 }-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 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 }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}-\frac{1}{6}=\frac{23}{24}: \\
-6 \left(x^2+\frac{x}{3}+\frac{1}{36}\right)+2 \left(y^2-\frac{3 y}{2}+\frac{9}{16}\right)=\fbox{$\frac{23}{24}$} \\
\end{array}
Step 8:
\begin{array}{l}
x^2+\frac{x}{3}+\frac{1}{36}=\left(x+\frac{1}{6}\right)^2: \\
-6 \fbox{$\left(x+\frac{1}{6}\right)^2$}+2 \left(y^2-\frac{3 y}{2}+\frac{9}{16}\right)=\frac{23}{24} \\
\end{array}
Step 9:
\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{} & -6 \left(x+\frac{1}{6}\right)^2+2 \fbox{$\left(y-\frac{3}{4}\right)^2$}=\frac{23}{24} \\
\end{array}
| khanacademy | amps |
Given the equation $5 x^2-x+8 y^2+6 y-7=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
8 y^2+6 y+5 x^2-x-7=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }7 \text{to }\text{both }\text{sides}: \\
8 y^2+6 y+5 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(5 x^2-x+\underline{\text{ }}\right)+\left(8 y^2+6 y+\underline{\text{ }}\right)=\underline{\text{ }}+7 \\
\end{array}
Step 4:
\begin{array}{l}
\left(5 x^2-x+\underline{\text{ }}\right)=5 \left(x^2-\frac{x}{5}+\underline{\text{ }}\right): \\
\fbox{$5 \left(x^2-\frac{x}{5}+\underline{\text{ }}\right)$}+\left(8 y^2+6 y+\underline{\text{ }}\right)=\underline{\text{ }}+7 \\
\end{array}
Step 5:
\begin{array}{l}
\left(8 y^2+6 y+\underline{\text{ }}\right)=8 \left(y^2+\frac{3 y}{4}+\underline{\text{ }}\right): \\
5 \left(x^2-\frac{x}{5}+\underline{\text{ }}\right)+\fbox{$8 \left(y^2+\frac{3 y}{4}+\underline{\text{ }}\right)$}=\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 }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 7:
\begin{array}{l}
7+\frac{1}{20}=\frac{141}{20}: \\
5 \left(x^2-\frac{x}{5}+\frac{1}{100}\right)+8 \left(y^2+\frac{3 y}{4}+\underline{\text{ }}\right)=\fbox{$\frac{141}{20}$} \\
\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 }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 9:
\begin{array}{l}
\frac{141}{20}+\frac{9}{8}=\frac{327}{40}: \\
5 \left(x^2-\frac{x}{5}+\frac{1}{100}\right)+8 \left(y^2+\frac{3 y}{4}+\frac{9}{64}\right)=\fbox{$\frac{327}{40}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{x}{5}+\frac{1}{100}=\left(x-\frac{1}{10}\right)^2: \\
5 \fbox{$\left(x-\frac{1}{10}\right)^2$}+8 \left(y^2+\frac{3 y}{4}+\frac{9}{64}\right)=\frac{327}{40} \\
\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{} & 5 \left(x-\frac{1}{10}\right)^2+8 \fbox{$\left(y+\frac{3}{8}\right)^2$}=\frac{327}{40} \\
\end{array}
| khanacademy | amps |
Given the equation $-6 x^2-x-5 y^2-5 y-5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-5 y^2-5 y-6 x^2-x-5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }5 \text{to }\text{both }\text{sides}: \\
-5 y^2-5 y-6 x^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(-6 x^2-x+\underline{\text{ }}\right)+\left(-5 y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}+5 \\
\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(-5 y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}+5 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-5 y^2-5 y+\underline{\text{ }}\right)=-5 \left(y^2+y+\underline{\text{ }}\right): \\
-6 \left(x^2+\frac{x}{6}+\underline{\text{ }}\right)+\fbox{$-5 \left(y^2+y+\underline{\text{ }}\right)$}=\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 }-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}
5-\frac{1}{24}=\frac{119}{24}: \\
-6 \left(x^2+\frac{x}{6}+\frac{1}{144}\right)-5 \left(y^2+y+\underline{\text{ }}\right)=\fbox{$\frac{119}{24}$} \\
\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 }-5 \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{-5}{4}=-\frac{5}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{119}{24}-\frac{5}{4}=\frac{89}{24}: \\
-6 \left(x^2+\frac{x}{6}+\frac{1}{144}\right)-5 \left(y^2+y+\frac{1}{4}\right)=\fbox{$\frac{89}{24}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{x}{6}+\frac{1}{144}=\left(x+\frac{1}{12}\right)^2: \\
-6 \fbox{$\left(x+\frac{1}{12}\right)^2$}-5 \left(y^2+y+\frac{1}{4}\right)=\frac{89}{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{1}{12}\right)^2-5 \fbox{$\left(y+\frac{1}{2}\right)^2$}=\frac{89}{24} \\
\end{array}
| khanacademy | amps |
Given the equation $-9 x^2+7 x+6 y^2-10 y+8=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
6 y^2-10 y-9 x^2+7 x+8=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }8 \text{from }\text{both }\text{sides}: \\
6 y^2-10 y-9 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(-9 x^2+7 x+\underline{\text{ }}\right)+\left(6 y^2-10 y+\underline{\text{ }}\right)=\underline{\text{ }}-8 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-9 x^2+7 x+\underline{\text{ }}\right)=-9 \left(x^2-\frac{7 x}{9}+\underline{\text{ }}\right): \\
\fbox{$-9 \left(x^2-\frac{7 x}{9}+\underline{\text{ }}\right)$}+\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): \\
-9 \left(x^2-\frac{7 x}{9}+\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 }-9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-7}{9}}{2}\right)^2=\frac{49}{324} \text{on }\text{the }\text{left }\text{and }-9\times \frac{49}{324}=-\frac{49}{36} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-8-\frac{49}{36}=-\frac{337}{36}: \\
-9 \left(x^2-\frac{7 x}{9}+\frac{49}{324}\right)+6 \left(y^2-\frac{5 y}{3}+\underline{\text{ }}\right)=\fbox{$-\frac{337}{36}$} \\
\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{337}{36}=-\frac{187}{36}: \\
-9 \left(x^2-\frac{7 x}{9}+\frac{49}{324}\right)+6 \left(y^2-\frac{5 y}{3}+\frac{25}{36}\right)=\fbox{$-\frac{187}{36}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{7 x}{9}+\frac{49}{324}=\left(x-\frac{7}{18}\right)^2: \\
-9 \fbox{$\left(x-\frac{7}{18}\right)^2$}+6 \left(y^2-\frac{5 y}{3}+\frac{25}{36}\right)=-\frac{187}{36} \\
\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{} & -9 \left(x-\frac{7}{18}\right)^2+6 \fbox{$\left(y-\frac{5}{6}\right)^2$}=-\frac{187}{36} \\
\end{array}
| khanacademy | amps |
Given the equation $7 x^2+6 x+8 y^2-3 y=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
8 y^2-3 y+7 x^2+6 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(7 x^2+6 x+\underline{\text{ }}\right)+\left(8 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}+0 \\
\end{array}
Step 3:
\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^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}+0 \\
\end{array}
Step 4:
\begin{array}{l}
\left(8 y^2-3 y+\underline{\text{ }}\right)=8 \left(y^2-\frac{3 y}{8}+\underline{\text{ }}\right): \\
7 \left(x^2+\frac{6 x}{7}+\underline{\text{ }}\right)+\fbox{$8 \left(y^2-\frac{3 y}{8}+\underline{\text{ }}\right)$}=\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 }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 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{-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}{7}+\frac{9}{32}=\frac{351}{224}: \\
7 \left(x^2+\frac{6 x}{7}+\frac{9}{49}\right)+8 \left(y^2-\frac{3 y}{8}+\frac{9}{256}\right)=\fbox{$\frac{351}{224}$} \\
\end{array}
Step 8:
\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{3 y}{8}+\frac{9}{256}\right)=\frac{351}{224} \\
\end{array}
Step 9:
\begin{array}{l}
y^2-\frac{3 y}{8}+\frac{9}{256}=\left(y-\frac{3}{16}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 7 \left(x+\frac{3}{7}\right)^2+8 \fbox{$\left(y-\frac{3}{16}\right)^2$}=\frac{351}{224} \\
\end{array}
| khanacademy | amps |
Given the equation $4 x^2+2 x+5 y^2+5 y-10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
5 y^2+5 y+4 x^2+2 x-10=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }\text{10 }\text{to }\text{both }\text{sides}: \\
5 y^2+5 y+4 x^2+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(4 x^2+2 x+\underline{\text{ }}\right)+\left(5 y^2+5 y+\underline{\text{ }}\right)=\underline{\text{ }}+10 \\
\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(5 y^2+5 y+\underline{\text{ }}\right)=\underline{\text{ }}+10 \\
\end{array}
Step 5:
\begin{array}{l}
\left(5 y^2+5 y+\underline{\text{ }}\right)=5 \left(y^2+y+\underline{\text{ }}\right): \\
4 \left(x^2+\frac{x}{2}+\underline{\text{ }}\right)+\fbox{$5 \left(y^2+y+\underline{\text{ }}\right)$}=\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 }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}
10+\frac{1}{4}=\frac{41}{4}: \\
4 \left(x^2+\frac{x}{2}+\frac{1}{16}\right)+5 \left(y^2+y+\underline{\text{ }}\right)=\fbox{$\frac{41}{4}$} \\
\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{1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{5}{4}=\frac{5}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{41}{4}+\frac{5}{4}=\frac{23}{2}: \\
4 \left(x^2+\frac{x}{2}+\frac{1}{16}\right)+5 \left(y^2+y+\frac{1}{4}\right)=\fbox{$\frac{23}{2}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{x}{2}+\frac{1}{16}=\left(x+\frac{1}{4}\right)^2: \\
4 \fbox{$\left(x+\frac{1}{4}\right)^2$}+5 \left(y^2+y+\frac{1}{4}\right)=\frac{23}{2} \\
\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{1}{4}\right)^2+5 \fbox{$\left(y+\frac{1}{2}\right)^2$}=\frac{23}{2} \\
\end{array}
| khanacademy | amps |
Given the equation $-6 x^2-6 x-y^2+8 y-1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-y^2+8 y-6 x^2-6 x-1=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }1 \text{to }\text{both }\text{sides}: \\
-y^2+8 y-6 x^2-6 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(-6 x^2-6 x+\underline{\text{ }}\right)+\left(-y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}+1 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-6 x^2-6 x+\underline{\text{ }}\right)=-6 \left(x^2+x+\underline{\text{ }}\right): \\
\fbox{$-6 \left(x^2+x+\underline{\text{ }}\right)$}+\left(-y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}+1 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-y^2+8 y+\underline{\text{ }}\right)=-\left(y^2-8 y+\underline{\text{ }}\right): \\
-6 \left(x^2+x+\underline{\text{ }}\right)+\fbox{$-\left(y^2-8 y+\underline{\text{ }}\right)$}=\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 }-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 7:
\begin{array}{l}
1-\frac{3}{2}=-\frac{1}{2}: \\
-6 \left(x^2+x+\frac{1}{4}\right)-\left(y^2-8 y+\underline{\text{ }}\right)=\fbox{$-\frac{1}{2}$} \\
\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 }-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 9:
\begin{array}{l}
-\frac{1}{2}-16=-\frac{33}{2}: \\
-6 \left(x^2+x+\frac{1}{4}\right)-\left(y^2-8 y+16\right)=\fbox{$-\frac{33}{2}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+x+\frac{1}{4}=\left(x+\frac{1}{2}\right)^2: \\
-6 \fbox{$\left(x+\frac{1}{2}\right)^2$}-\left(y^2-8 y+16\right)=-\frac{33}{2} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-8 y+16=(y-4)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -6 \left(x+\frac{1}{2}\right)^2-\fbox{$(y-4)^2$}=-\frac{33}{2} \\
\end{array}
| khanacademy | amps |
Given the equation $7 x^2-x-8 y^2-3 y+5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-8 y^2-3 y+7 x^2-x+5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }5 \text{from }\text{both }\text{sides}: \\
-8 y^2-3 y+7 x^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(7 x^2-x+\underline{\text{ }}\right)+\left(-8 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\end{array}
Step 4:
\begin{array}{l}
\left(7 x^2-x+\underline{\text{ }}\right)=7 \left(x^2-\frac{x}{7}+\underline{\text{ }}\right): \\
\fbox{$7 \left(x^2-\frac{x}{7}+\underline{\text{ }}\right)$}+\left(-8 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-8 y^2-3 y+\underline{\text{ }}\right)=-8 \left(y^2+\frac{3 y}{8}+\underline{\text{ }}\right): \\
7 \left(x^2-\frac{x}{7}+\underline{\text{ }}\right)+\fbox{$-8 \left(y^2+\frac{3 y}{8}+\underline{\text{ }}\right)$}=\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{-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}-5=-\frac{139}{28}: \\
7 \left(x^2-\frac{x}{7}+\frac{1}{196}\right)-8 \left(y^2+\frac{3 y}{8}+\underline{\text{ }}\right)=\fbox{$-\frac{139}{28}$} \\
\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 }-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 9:
\begin{array}{l}
-\frac{139}{28}-\frac{9}{32}=-\frac{1175}{224}: \\
7 \left(x^2-\frac{x}{7}+\frac{1}{196}\right)-8 \left(y^2+\frac{3 y}{8}+\frac{9}{256}\right)=\fbox{$-\frac{1175}{224}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{x}{7}+\frac{1}{196}=\left(x-\frac{1}{14}\right)^2: \\
7 \fbox{$\left(x-\frac{1}{14}\right)^2$}-8 \left(y^2+\frac{3 y}{8}+\frac{9}{256}\right)=-\frac{1175}{224} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{3 y}{8}+\frac{9}{256}=\left(y+\frac{3}{16}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 7 \left(x-\frac{1}{14}\right)^2-8 \fbox{$\left(y+\frac{3}{16}\right)^2$}=-\frac{1175}{224} \\
\end{array}
| khanacademy | amps |
Given the equation $-3 x^2-8 x+10 y^2-8 y+6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
10 y^2-8 y-3 x^2-8 x+6=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }6 \text{from }\text{both }\text{sides}: \\
10 y^2-8 y-3 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(-3 x^2-8 x+\underline{\text{ }}\right)+\left(10 y^2-8 y+\underline{\text{ }}\right)=\underline{\text{ }}-6 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-3 x^2-8 x+\underline{\text{ }}\right)=-3 \left(x^2+\frac{8 x}{3}+\underline{\text{ }}\right): \\
\fbox{$-3 \left(x^2+\frac{8 x}{3}+\underline{\text{ }}\right)$}+\left(10 y^2-8 y+\underline{\text{ }}\right)=\underline{\text{ }}-6 \\
\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): \\
-3 \left(x^2+\frac{8 x}{3}+\underline{\text{ }}\right)+\fbox{$10 \left(y^2-\frac{4 y}{5}+\underline{\text{ }}\right)$}=\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 }-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 7:
\begin{array}{l}
-6-\frac{16}{3}=-\frac{34}{3}: \\
-3 \left(x^2+\frac{8 x}{3}+\frac{16}{9}\right)+10 \left(y^2-\frac{4 y}{5}+\underline{\text{ }}\right)=\fbox{$-\frac{34}{3}$} \\
\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 }\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{8}{5}-\frac{34}{3}=-\frac{146}{15}: \\
-3 \left(x^2+\frac{8 x}{3}+\frac{16}{9}\right)+10 \left(y^2-\frac{4 y}{5}+\frac{4}{25}\right)=\fbox{$-\frac{146}{15}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{8 x}{3}+\frac{16}{9}=\left(x+\frac{4}{3}\right)^2: \\
-3 \fbox{$\left(x+\frac{4}{3}\right)^2$}+10 \left(y^2-\frac{4 y}{5}+\frac{4}{25}\right)=-\frac{146}{15} \\
\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{} & -3 \left(x+\frac{4}{3}\right)^2+\text{10 }\fbox{$\left(y-\frac{2}{5}\right)^2$}=-\frac{146}{15} \\
\end{array}
| khanacademy | amps |
Given the equation $-x^2+7 x-9 y^2+8 y-2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-9 y^2+8 y-x^2+7 x-2=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }2 \text{to }\text{both }\text{sides}: \\
-9 y^2+8 y-x^2+7 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+7 x+\underline{\text{ }}\right)+\left(-9 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}+2 \\
\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+8 y+\underline{\text{ }}\right)=\underline{\text{ }}+2 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-9 y^2+8 y+\underline{\text{ }}\right)=-9 \left(y^2-\frac{8 y}{9}+\underline{\text{ }}\right): \\
-\left(x^2-7 x+\underline{\text{ }}\right)+\fbox{$-9 \left(y^2-\frac{8 y}{9}+\underline{\text{ }}\right)$}=\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{-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}
2-\frac{49}{4}=-\frac{41}{4}: \\
-\left(x^2-7 x+\frac{49}{4}\right)-9 \left(y^2-\frac{8 y}{9}+\underline{\text{ }}\right)=\fbox{$-\frac{41}{4}$} \\
\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{-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 9:
\begin{array}{l}
-\frac{41}{4}-\frac{16}{9}=-\frac{433}{36}: \\
-\left(x^2-7 x+\frac{49}{4}\right)-9 \left(y^2-\frac{8 y}{9}+\frac{16}{81}\right)=\fbox{$-\frac{433}{36}$} \\
\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{8 y}{9}+\frac{16}{81}\right)=-\frac{433}{36} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{8 y}{9}+\frac{16}{81}=\left(y-\frac{4}{9}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -\left(x-\frac{7}{2}\right)^2-9 \fbox{$\left(y-\frac{4}{9}\right)^2$}=-\frac{433}{36} \\
\end{array}
| khanacademy | amps |
Given the equation $5 x^2+5 x-5 y^2+2 y-9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-5 y^2+2 y+5 x^2+5 x-9=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }9 \text{to }\text{both }\text{sides}: \\
-5 y^2+2 y+5 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(5 x^2+5 x+\underline{\text{ }}\right)+\left(-5 y^2+2 y+\underline{\text{ }}\right)=\underline{\text{ }}+9 \\
\end{array}
Step 4:
\begin{array}{l}
\left(5 x^2+5 x+\underline{\text{ }}\right)=5 \left(x^2+x+\underline{\text{ }}\right): \\
\fbox{$5 \left(x^2+x+\underline{\text{ }}\right)$}+\left(-5 y^2+2 y+\underline{\text{ }}\right)=\underline{\text{ }}+9 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-5 y^2+2 y+\underline{\text{ }}\right)=-5 \left(y^2-\frac{2 y}{5}+\underline{\text{ }}\right): \\
5 \left(x^2+x+\underline{\text{ }}\right)+\fbox{$-5 \left(y^2-\frac{2 y}{5}+\underline{\text{ }}\right)$}=\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 }5 \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{5}{4}=\frac{5}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
9+\frac{5}{4}=\frac{41}{4}: \\
5 \left(x^2+x+\frac{1}{4}\right)-5 \left(y^2-\frac{2 y}{5}+\underline{\text{ }}\right)=\fbox{$\frac{41}{4}$} \\
\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{-2}{5}}{2}\right)^2=\frac{1}{25} \text{on }\text{the }\text{left }\text{and }\frac{-5}{25}=-\frac{1}{5} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{41}{4}-\frac{1}{5}=\frac{201}{20}: \\
5 \left(x^2+x+\frac{1}{4}\right)-5 \left(y^2-\frac{2 y}{5}+\frac{1}{25}\right)=\fbox{$\frac{201}{20}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+x+\frac{1}{4}=\left(x+\frac{1}{2}\right)^2: \\
5 \fbox{$\left(x+\frac{1}{2}\right)^2$}-5 \left(y^2-\frac{2 y}{5}+\frac{1}{25}\right)=\frac{201}{20} \\
\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{} & 5 \left(x+\frac{1}{2}\right)^2-5 \fbox{$\left(y-\frac{1}{5}\right)^2$}=\frac{201}{20} \\
\end{array}
| khanacademy | amps |
Given the equation $3 x^2+4 x-9 y^2+10 y=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-9 y^2+10 y+3 x^2+4 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(3 x^2+4 x+\underline{\text{ }}\right)+\left(-9 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}+0 \\
\end{array}
Step 3:
\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(-9 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}+0 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-9 y^2+10 y+\underline{\text{ }}\right)=-9 \left(y^2-\frac{10 y}{9}+\underline{\text{ }}\right): \\
3 \left(x^2+\frac{4 x}{3}+\underline{\text{ }}\right)+\fbox{$-9 \left(y^2-\frac{10 y}{9}+\underline{\text{ }}\right)$}=\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 }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 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{-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}
\frac{4}{3}-\frac{25}{9}=-\frac{13}{9}: \\
3 \left(x^2+\frac{4 x}{3}+\frac{4}{9}\right)-9 \left(y^2-\frac{10 y}{9}+\frac{25}{81}\right)=\fbox{$-\frac{13}{9}$} \\
\end{array}
Step 8:
\begin{array}{l}
x^2+\frac{4 x}{3}+\frac{4}{9}=\left(x+\frac{2}{3}\right)^2: \\
3 \fbox{$\left(x+\frac{2}{3}\right)^2$}-9 \left(y^2-\frac{10 y}{9}+\frac{25}{81}\right)=-\frac{13}{9} \\
\end{array}
Step 9:
\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{} & 3 \left(x+\frac{2}{3}\right)^2-9 \fbox{$\left(y-\frac{5}{9}\right)^2$}=-\frac{13}{9} \\
\end{array}
| khanacademy | amps |
Given the equation $-8 x^2+2 x+10 y^2-9 y+6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
10 y^2-9 y-8 x^2+2 x+6=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }6 \text{from }\text{both }\text{sides}: \\
10 y^2-9 y-8 x^2+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(-8 x^2+2 x+\underline{\text{ }}\right)+\left(10 y^2-9 y+\underline{\text{ }}\right)=\underline{\text{ }}-6 \\
\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(10 y^2-9 y+\underline{\text{ }}\right)=\underline{\text{ }}-6 \\
\end{array}
Step 5:
\begin{array}{l}
\left(10 y^2-9 y+\underline{\text{ }}\right)=10 \left(y^2-\frac{9 y}{10}+\underline{\text{ }}\right): \\
-8 \left(x^2-\frac{x}{4}+\underline{\text{ }}\right)+\fbox{$10 \left(y^2-\frac{9 y}{10}+\underline{\text{ }}\right)$}=\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{-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}
-6-\frac{1}{8}=-\frac{49}{8}: \\
-8 \left(x^2-\frac{x}{4}+\frac{1}{64}\right)+10 \left(y^2-\frac{9 y}{10}+\underline{\text{ }}\right)=\fbox{$-\frac{49}{8}$} \\
\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{-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 9:
\begin{array}{l}
\frac{81}{40}-\frac{49}{8}=-\frac{41}{10}: \\
-8 \left(x^2-\frac{x}{4}+\frac{1}{64}\right)+10 \left(y^2-\frac{9 y}{10}+\frac{81}{400}\right)=\fbox{$-\frac{41}{10}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{x}{4}+\frac{1}{64}=\left(x-\frac{1}{8}\right)^2: \\
-8 \fbox{$\left(x-\frac{1}{8}\right)^2$}+10 \left(y^2-\frac{9 y}{10}+\frac{81}{400}\right)=-\frac{41}{10} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{9 y}{10}+\frac{81}{400}=\left(y-\frac{9}{20}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -8 \left(x-\frac{1}{8}\right)^2+\text{10 }\fbox{$\left(y-\frac{9}{20}\right)^2$}=-\frac{41}{10} \\
\end{array}
| khanacademy | amps |
Given the equation $-6 x^2+3 x+7 y^2+y+5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
7 y^2+y-6 x^2+3 x+5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }5 \text{from }\text{both }\text{sides}: \\
7 y^2+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(7 y^2+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(7 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\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): \\
-6 \left(x^2-\frac{x}{2}+\underline{\text{ }}\right)+\fbox{$7 \left(y^2+\frac{y}{7}+\underline{\text{ }}\right)$}=\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}
-5-\frac{3}{8}=-\frac{43}{8}: \\
-6 \left(x^2-\frac{x}{2}+\frac{1}{16}\right)+7 \left(y^2+\frac{y}{7}+\underline{\text{ }}\right)=\fbox{$-\frac{43}{8}$} \\
\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{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{1}{28}-\frac{43}{8}=-\frac{299}{56}: \\
-6 \left(x^2-\frac{x}{2}+\frac{1}{16}\right)+7 \left(y^2+\frac{y}{7}+\frac{1}{196}\right)=\fbox{$-\frac{299}{56}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{x}{2}+\frac{1}{16}=\left(x-\frac{1}{4}\right)^2: \\
-6 \fbox{$\left(x-\frac{1}{4}\right)^2$}+7 \left(y^2+\frac{y}{7}+\frac{1}{196}\right)=-\frac{299}{56} \\
\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{} & -6 \left(x-\frac{1}{4}\right)^2+7 \fbox{$\left(y+\frac{1}{14}\right)^2$}=-\frac{299}{56} \\
\end{array}
| khanacademy | amps |
Given the equation $-7 x^2+4 x+2 y^2+6 y+10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
2 y^2+6 y-7 x^2+4 x+10=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }\text{10 }\text{from }\text{both }\text{sides}: \\
2 y^2+6 y-7 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(-7 x^2+4 x+\underline{\text{ }}\right)+\left(2 y^2+6 y+\underline{\text{ }}\right)=\underline{\text{ }}-10 \\
\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(2 y^2+6 y+\underline{\text{ }}\right)=\underline{\text{ }}-10 \\
\end{array}
Step 5:
\begin{array}{l}
\left(2 y^2+6 y+\underline{\text{ }}\right)=2 \left(y^2+3 y+\underline{\text{ }}\right): \\
-7 \left(x^2-\frac{4 x}{7}+\underline{\text{ }}\right)+\fbox{$2 \left(y^2+3 y+\underline{\text{ }}\right)$}=\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 }-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}
-10-\frac{4}{7}=-\frac{74}{7}: \\
-7 \left(x^2-\frac{4 x}{7}+\frac{4}{49}\right)+2 \left(y^2+3 y+\underline{\text{ }}\right)=\fbox{$-\frac{74}{7}$} \\
\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{3}{2}\right)^2=\frac{9}{4} \text{on }\text{the }\text{left }\text{and }2\times \frac{9}{4}=\frac{9}{2} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{9}{2}-\frac{74}{7}=-\frac{85}{14}: \\
-7 \left(x^2-\frac{4 x}{7}+\frac{4}{49}\right)+2 \left(y^2+3 y+\frac{9}{4}\right)=\fbox{$-\frac{85}{14}$} \\
\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{$\left(x-\frac{2}{7}\right)^2$}+2 \left(y^2+3 y+\frac{9}{4}\right)=-\frac{85}{14} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+3 y+\frac{9}{4}=\left(y+\frac{3}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -7 \left(x-\frac{2}{7}\right)^2+2 \fbox{$\left(y+\frac{3}{2}\right)^2$}=-\frac{85}{14} \\
\end{array}
| khanacademy | amps |
Given the equation $10 x^2+2 x-3 y^2+10 y-1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-3 y^2+10 y+10 x^2+2 x-1=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }1 \text{to }\text{both }\text{sides}: \\
-3 y^2+10 y+10 x^2+2 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(10 x^2+2 x+\underline{\text{ }}\right)+\left(-3 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}+1 \\
\end{array}
Step 4:
\begin{array}{l}
\left(10 x^2+2 x+\underline{\text{ }}\right)=10 \left(x^2+\frac{x}{5}+\underline{\text{ }}\right): \\
\fbox{$10 \left(x^2+\frac{x}{5}+\underline{\text{ }}\right)$}+\left(-3 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}+1 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-3 y^2+10 y+\underline{\text{ }}\right)=-3 \left(y^2-\frac{10 y}{3}+\underline{\text{ }}\right): \\
10 \left(x^2+\frac{x}{5}+\underline{\text{ }}\right)+\fbox{$-3 \left(y^2-\frac{10 y}{3}+\underline{\text{ }}\right)$}=\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 }\text{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 7:
\begin{array}{l}
1+\frac{1}{10}=\frac{11}{10}: \\
10 \left(x^2+\frac{x}{5}+\frac{1}{100}\right)-3 \left(y^2-\frac{10 y}{3}+\underline{\text{ }}\right)=\fbox{$\frac{11}{10}$} \\
\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 }-3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-10}{3}}{2}\right)^2=\frac{25}{9} \text{on }\text{the }\text{left }\text{and }-3\times \frac{25}{9}=-\frac{25}{3} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{11}{10}-\frac{25}{3}=-\frac{217}{30}: \\
10 \left(x^2+\frac{x}{5}+\frac{1}{100}\right)-3 \left(y^2-\frac{10 y}{3}+\frac{25}{9}\right)=\fbox{$-\frac{217}{30}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{x}{5}+\frac{1}{100}=\left(x+\frac{1}{10}\right)^2: \\
\text{10 }\fbox{$\left(x+\frac{1}{10}\right)^2$}-3 \left(y^2-\frac{10 y}{3}+\frac{25}{9}\right)=-\frac{217}{30} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{10 y}{3}+\frac{25}{9}=\left(y-\frac{5}{3}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 10 \left(x+\frac{1}{10}\right)^2-3 \fbox{$\left(y-\frac{5}{3}\right)^2$}=-\frac{217}{30} \\
\end{array}
| khanacademy | amps |
Given the equation $4 x^2-5 x-5 y^2-5 y-9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-5 y^2-5 y+4 x^2-5 x-9=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }9 \text{to }\text{both }\text{sides}: \\
-5 y^2-5 y+4 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(4 x^2-5 x+\underline{\text{ }}\right)+\left(-5 y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}+9 \\
\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(-5 y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}+9 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-5 y^2-5 y+\underline{\text{ }}\right)=-5 \left(y^2+y+\underline{\text{ }}\right): \\
4 \left(x^2-\frac{5 x}{4}+\underline{\text{ }}\right)+\fbox{$-5 \left(y^2+y+\underline{\text{ }}\right)$}=\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{-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}
9+\frac{25}{16}=\frac{169}{16}: \\
4 \left(x^2-\frac{5 x}{4}+\frac{25}{64}\right)-5 \left(y^2+y+\underline{\text{ }}\right)=\fbox{$\frac{169}{16}$} \\
\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 }-5 \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{-5}{4}=-\frac{5}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{169}{16}-\frac{5}{4}=\frac{149}{16}: \\
4 \left(x^2-\frac{5 x}{4}+\frac{25}{64}\right)-5 \left(y^2+y+\frac{1}{4}\right)=\fbox{$\frac{149}{16}$} \\
\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{$\left(x-\frac{5}{8}\right)^2$}-5 \left(y^2+y+\frac{1}{4}\right)=\frac{149}{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{5}{8}\right)^2-5 \fbox{$\left(y+\frac{1}{2}\right)^2$}=\frac{149}{16} \\
\end{array}
| khanacademy | amps |
Given the equation $-10 x^2-4 x+2 y^2-9 y+7=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
2 y^2-9 y-10 x^2-4 x+7=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }7 \text{from }\text{both }\text{sides}: \\
2 y^2-9 y-10 x^2-4 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^2-4 x+\underline{\text{ }}\right)+\left(2 y^2-9 y+\underline{\text{ }}\right)=\underline{\text{ }}-7 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-10 x^2-4 x+\underline{\text{ }}\right)=-10 \left(x^2+\frac{2 x}{5}+\underline{\text{ }}\right): \\
\fbox{$-10 \left(x^2+\frac{2 x}{5}+\underline{\text{ }}\right)$}+\left(2 y^2-9 y+\underline{\text{ }}\right)=\underline{\text{ }}-7 \\
\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): \\
-10 \left(x^2+\frac{2 x}{5}+\underline{\text{ }}\right)+\fbox{$2 \left(y^2-\frac{9 y}{2}+\underline{\text{ }}\right)$}=\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 }-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 7:
\begin{array}{l}
-7-\frac{2}{5}=-\frac{37}{5}: \\
-10 \left(x^2+\frac{2 x}{5}+\frac{1}{25}\right)+2 \left(y^2-\frac{9 y}{2}+\underline{\text{ }}\right)=\fbox{$-\frac{37}{5}$} \\
\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 }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}{5}=\frac{109}{40}: \\
-10 \left(x^2+\frac{2 x}{5}+\frac{1}{25}\right)+2 \left(y^2-\frac{9 y}{2}+\frac{81}{16}\right)=\fbox{$\frac{109}{40}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{2 x}{5}+\frac{1}{25}=\left(x+\frac{1}{5}\right)^2: \\
-10 \fbox{$\left(x+\frac{1}{5}\right)^2$}+2 \left(y^2-\frac{9 y}{2}+\frac{81}{16}\right)=\frac{109}{40} \\
\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{} & -10 \left(x+\frac{1}{5}\right)^2+2 \fbox{$\left(y-\frac{9}{4}\right)^2$}=\frac{109}{40} \\
\end{array}
| khanacademy | amps |
Given the equation $-8 x^2+7 x+3 y^2+4 y-10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
3 y^2+4 y-8 x^2+7 x-10=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }\text{10 }\text{to }\text{both }\text{sides}: \\
3 y^2+4 y-8 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(-8 x^2+7 x+\underline{\text{ }}\right)+\left(3 y^2+4 y+\underline{\text{ }}\right)=\underline{\text{ }}+10 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-8 x^2+7 x+\underline{\text{ }}\right)=-8 \left(x^2-\frac{7 x}{8}+\underline{\text{ }}\right): \\
\fbox{$-8 \left(x^2-\frac{7 x}{8}+\underline{\text{ }}\right)$}+\left(3 y^2+4 y+\underline{\text{ }}\right)=\underline{\text{ }}+10 \\
\end{array}
Step 5:
\begin{array}{l}
\left(3 y^2+4 y+\underline{\text{ }}\right)=3 \left(y^2+\frac{4 y}{3}+\underline{\text{ }}\right): \\
-8 \left(x^2-\frac{7 x}{8}+\underline{\text{ }}\right)+\fbox{$3 \left(y^2+\frac{4 y}{3}+\underline{\text{ }}\right)$}=\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 }-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 7:
\begin{array}{l}
10-\frac{49}{32}=\frac{271}{32}: \\
-8 \left(x^2-\frac{7 x}{8}+\frac{49}{256}\right)+3 \left(y^2+\frac{4 y}{3}+\underline{\text{ }}\right)=\fbox{$\frac{271}{32}$} \\
\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 }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 9:
\begin{array}{l}
\frac{271}{32}+\frac{4}{3}=\frac{941}{96}: \\
-8 \left(x^2-\frac{7 x}{8}+\frac{49}{256}\right)+3 \left(y^2+\frac{4 y}{3}+\frac{4}{9}\right)=\fbox{$\frac{941}{96}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{7 x}{8}+\frac{49}{256}=\left(x-\frac{7}{16}\right)^2: \\
-8 \fbox{$\left(x-\frac{7}{16}\right)^2$}+3 \left(y^2+\frac{4 y}{3}+\frac{4}{9}\right)=\frac{941}{96} \\
\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{} & -8 \left(x-\frac{7}{16}\right)^2+3 \fbox{$\left(y+\frac{2}{3}\right)^2$}=\frac{941}{96} \\
\end{array}
| khanacademy | amps |
Given the equation $-2 x^2+10 x-8 y-2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-2 x^2+10 x+(-8 y-2)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }-8 y-2 x^2+10 x-2 \text{from }\text{both }\text{sides}: \\
2 x^2-10 x+(8 y+2)=0 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Subtract }8 y+2 \text{from }\text{both }\text{sides}: \\
2 x^2-10 x=-8 y-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(2 x^2-10 x+\underline{\text{ }}\right)=(-8 y-2)+\underline{\text{ }} \\
\end{array}
Step 5:
\begin{array}{l}
\left(2 x^2-10 x+\underline{\text{ }}\right)=2 \left(x^2-5 x+\underline{\text{ }}\right): \\
\fbox{$2 \left(x^2-5 x+\underline{\text{ }}\right)$}=(-8 y-2)+\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{-5}{2}\right)^2=\frac{25}{4} \text{on }\text{the }\text{left }\text{and }2\times \frac{25}{4}=\frac{25}{2} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
(-8 y-2)+\frac{25}{2}=\frac{21}{2}-8 y: \\
2 \left(x^2-5 x+\frac{25}{4}\right)=\fbox{$\frac{21}{2}-8 y$} \\
\end{array}
Step 8:
\begin{array}{l}
x^2-5 x+\frac{25}{4}=\left(x-\frac{5}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 2 \fbox{$\left(x-\frac{5}{2}\right)^2$}=\frac{21}{2}-8 y \\
\end{array}
| khanacademy | amps |
Given the equation $8 x^2+5 x+9 y^2-6 y-6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
9 y^2-6 y+8 x^2+5 x-6=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }6 \text{to }\text{both }\text{sides}: \\
9 y^2-6 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(9 y^2-6 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(9 y^2-6 y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\end{array}
Step 5:
\begin{array}{l}
\left(9 y^2-6 y+\underline{\text{ }}\right)=9 \left(y^2-\frac{2 y}{3}+\underline{\text{ }}\right): \\
8 \left(x^2+\frac{5 x}{8}+\underline{\text{ }}\right)+\fbox{$9 \left(y^2-\frac{2 y}{3}+\underline{\text{ }}\right)$}=\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 }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{217}{32}: \\
8 \left(x^2+\frac{5 x}{8}+\frac{25}{256}\right)+9 \left(y^2-\frac{2 y}{3}+\underline{\text{ }}\right)=\fbox{$\frac{217}{32}$} \\
\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 }9 \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{9}{9}=1 \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{217}{32}+1=\frac{249}{32}: \\
8 \left(x^2+\frac{5 x}{8}+\frac{25}{256}\right)+9 \left(y^2-\frac{2 y}{3}+\frac{1}{9}\right)=\fbox{$\frac{249}{32}$} \\
\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{$\left(x+\frac{5}{16}\right)^2$}+9 \left(y^2-\frac{2 y}{3}+\frac{1}{9}\right)=\frac{249}{32} \\
\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{} & 8 \left(x+\frac{5}{16}\right)^2+9 \fbox{$\left(y-\frac{1}{3}\right)^2$}=\frac{249}{32} \\
\end{array}
| khanacademy | amps |
Given the equation $3 x^2+2 x+9 y^2-7 y+3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
9 y^2-7 y+3 x^2+2 x+3=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }3 \text{from }\text{both }\text{sides}: \\
9 y^2-7 y+3 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(3 x^2+2 x+\underline{\text{ }}\right)+\left(9 y^2-7 y+\underline{\text{ }}\right)=\underline{\text{ }}-3 \\
\end{array}
Step 4:
\begin{array}{l}
\left(3 x^2+2 x+\underline{\text{ }}\right)=3 \left(x^2+\frac{2 x}{3}+\underline{\text{ }}\right): \\
\fbox{$3 \left(x^2+\frac{2 x}{3}+\underline{\text{ }}\right)$}+\left(9 y^2-7 y+\underline{\text{ }}\right)=\underline{\text{ }}-3 \\
\end{array}
Step 5:
\begin{array}{l}
\left(9 y^2-7 y+\underline{\text{ }}\right)=9 \left(y^2-\frac{7 y}{9}+\underline{\text{ }}\right): \\
3 \left(x^2+\frac{2 x}{3}+\underline{\text{ }}\right)+\fbox{$9 \left(y^2-\frac{7 y}{9}+\underline{\text{ }}\right)$}=\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 }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 7:
\begin{array}{l}
\frac{1}{3}-3=-\frac{8}{3}: \\
3 \left(x^2+\frac{2 x}{3}+\frac{1}{9}\right)+9 \left(y^2-\frac{7 y}{9}+\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 }9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-7}{9}}{2}\right)^2=\frac{49}{324} \text{on }\text{the }\text{left }\text{and }9\times \frac{49}{324}=\frac{49}{36} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{49}{36}-\frac{8}{3}=-\frac{47}{36}: \\
3 \left(x^2+\frac{2 x}{3}+\frac{1}{9}\right)+9 \left(y^2-\frac{7 y}{9}+\frac{49}{324}\right)=\fbox{$-\frac{47}{36}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{2 x}{3}+\frac{1}{9}=\left(x+\frac{1}{3}\right)^2: \\
3 \fbox{$\left(x+\frac{1}{3}\right)^2$}+9 \left(y^2-\frac{7 y}{9}+\frac{49}{324}\right)=-\frac{47}{36} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{7 y}{9}+\frac{49}{324}=\left(y-\frac{7}{18}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 3 \left(x+\frac{1}{3}\right)^2+9 \fbox{$\left(y-\frac{7}{18}\right)^2$}=-\frac{47}{36} \\
\end{array}
| khanacademy | amps |
Given the equation $8 x^2+7 x+2 y^2+5 y+1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
2 y^2+5 y+8 x^2+7 x+1=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }1 \text{from }\text{both }\text{sides}: \\
2 y^2+5 y+8 x^2+7 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+7 x+\underline{\text{ }}\right)+\left(2 y^2+5 y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\end{array}
Step 4:
\begin{array}{l}
\left(8 x^2+7 x+\underline{\text{ }}\right)=8 \left(x^2+\frac{7 x}{8}+\underline{\text{ }}\right): \\
\fbox{$8 \left(x^2+\frac{7 x}{8}+\underline{\text{ }}\right)$}+\left(2 y^2+5 y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\end{array}
Step 5:
\begin{array}{l}
\left(2 y^2+5 y+\underline{\text{ }}\right)=2 \left(y^2+\frac{5 y}{2}+\underline{\text{ }}\right): \\
8 \left(x^2+\frac{7 x}{8}+\underline{\text{ }}\right)+\fbox{$2 \left(y^2+\frac{5 y}{2}+\underline{\text{ }}\right)$}=\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{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 7:
\begin{array}{l}
\frac{49}{32}-1=\frac{17}{32}: \\
8 \left(x^2+\frac{7 x}{8}+\frac{49}{256}\right)+2 \left(y^2+\frac{5 y}{2}+\underline{\text{ }}\right)=\fbox{$\frac{17}{32}$} \\
\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 }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 9:
\begin{array}{l}
\frac{17}{32}+\frac{25}{8}=\frac{117}{32}: \\
8 \left(x^2+\frac{7 x}{8}+\frac{49}{256}\right)+2 \left(y^2+\frac{5 y}{2}+\frac{25}{16}\right)=\fbox{$\frac{117}{32}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{7 x}{8}+\frac{49}{256}=\left(x+\frac{7}{16}\right)^2: \\
8 \fbox{$\left(x+\frac{7}{16}\right)^2$}+2 \left(y^2+\frac{5 y}{2}+\frac{25}{16}\right)=\frac{117}{32} \\
\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{} & 8 \left(x+\frac{7}{16}\right)^2+2 \fbox{$\left(y+\frac{5}{4}\right)^2$}=\frac{117}{32} \\
\end{array}
| khanacademy | amps |
Given the equation $6 x^2+8 x+y^2-6 y-10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
y^2-6 y+6 x^2+8 x-10=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }\text{10 }\text{to }\text{both }\text{sides}: \\
y^2-6 y+6 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(6 x^2+8 x+\underline{\text{ }}\right)+\left(y^2-6 y+\underline{\text{ }}\right)=\underline{\text{ }}+10 \\
\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^2-6 y+\underline{\text{ }}\right)=\underline{\text{ }}+10 \\
\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 }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 6:
\begin{array}{l}
10+\frac{8}{3}=\frac{38}{3}: \\
6 \left(x^2+\frac{4 x}{3}+\frac{4}{9}\right)+\left(y^2-6 y+\underline{\text{ }}\right)=\fbox{$\frac{38}{3}$} \\
\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}
\frac{38}{3}+9=\frac{65}{3}: \\
6 \left(x^2+\frac{4 x}{3}+\frac{4}{9}\right)+\left(y^2-6 y+9\right)=\fbox{$\frac{65}{3}$} \\
\end{array}
Step 9:
\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^2-6 y+9\right)=\frac{65}{3} \\
\end{array}
Step 10:
\begin{array}{l}
y^2-6 y+9=(y-3)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 6 \left(x+\frac{2}{3}\right)^2+\fbox{$(y-3)^2$}=\frac{65}{3} \\
\end{array}
| khanacademy | amps |
Given the equation $-8 x^2-3 x-4 y^2-y-8=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-4 y^2-y-8 x^2-3 x-8=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }8 \text{to }\text{both }\text{sides}: \\
-4 y^2-y-8 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(-8 x^2-3 x+\underline{\text{ }}\right)+\left(-4 y^2-y+\underline{\text{ }}\right)=\underline{\text{ }}+8 \\
\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(-4 y^2-y+\underline{\text{ }}\right)=\underline{\text{ }}+8 \\
\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): \\
-8 \left(x^2+\frac{3 x}{8}+\underline{\text{ }}\right)+\fbox{$-4 \left(y^2+\frac{y}{4}+\underline{\text{ }}\right)$}=\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 }-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}
8-\frac{9}{32}=\frac{247}{32}: \\
-8 \left(x^2+\frac{3 x}{8}+\frac{9}{256}\right)-4 \left(y^2+\frac{y}{4}+\underline{\text{ }}\right)=\fbox{$\frac{247}{32}$} \\
\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}{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{247}{32}-\frac{1}{16}=\frac{245}{32}: \\
-8 \left(x^2+\frac{3 x}{8}+\frac{9}{256}\right)-4 \left(y^2+\frac{y}{4}+\frac{1}{64}\right)=\fbox{$\frac{245}{32}$} \\
\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{$\left(x+\frac{3}{16}\right)^2$}-4 \left(y^2+\frac{y}{4}+\frac{1}{64}\right)=\frac{245}{32} \\
\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{} & -8 \left(x+\frac{3}{16}\right)^2-4 \fbox{$\left(y+\frac{1}{8}\right)^2$}=\frac{245}{32} \\
\end{array}
| khanacademy | amps |
Given the equation $8 x^2+3 x-3 y^2+2 y-3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-3 y^2+2 y+8 x^2+3 x-3=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }3 \text{to }\text{both }\text{sides}: \\
-3 y^2+2 y+8 x^2+3 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+3 x+\underline{\text{ }}\right)+\left(-3 y^2+2 y+\underline{\text{ }}\right)=\underline{\text{ }}+3 \\
\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(-3 y^2+2 y+\underline{\text{ }}\right)=\underline{\text{ }}+3 \\
\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): \\
8 \left(x^2+\frac{3 x}{8}+\underline{\text{ }}\right)+\fbox{$-3 \left(y^2-\frac{2 y}{3}+\underline{\text{ }}\right)$}=\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 }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}
3+\frac{9}{32}=\frac{105}{32}: \\
8 \left(x^2+\frac{3 x}{8}+\frac{9}{256}\right)-3 \left(y^2-\frac{2 y}{3}+\underline{\text{ }}\right)=\fbox{$\frac{105}{32}$} \\
\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 }-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{105}{32}-\frac{1}{3}=\frac{283}{96}: \\
8 \left(x^2+\frac{3 x}{8}+\frac{9}{256}\right)-3 \left(y^2-\frac{2 y}{3}+\frac{1}{9}\right)=\fbox{$\frac{283}{96}$} \\
\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{$\left(x+\frac{3}{16}\right)^2$}-3 \left(y^2-\frac{2 y}{3}+\frac{1}{9}\right)=\frac{283}{96} \\
\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{} & 8 \left(x+\frac{3}{16}\right)^2-3 \fbox{$\left(y-\frac{1}{3}\right)^2$}=\frac{283}{96} \\
\end{array}
| khanacademy | amps |
Given the equation $6 x^2-3 x+9 y^2-7 y+9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
9 y^2-7 y+6 x^2-3 x+9=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }9 \text{from }\text{both }\text{sides}: \\
9 y^2-7 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(9 y^2-7 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(9 y^2-7 y+\underline{\text{ }}\right)=\underline{\text{ }}-9 \\
\end{array}
Step 5:
\begin{array}{l}
\left(9 y^2-7 y+\underline{\text{ }}\right)=9 \left(y^2-\frac{7 y}{9}+\underline{\text{ }}\right): \\
6 \left(x^2-\frac{x}{2}+\underline{\text{ }}\right)+\fbox{$9 \left(y^2-\frac{7 y}{9}+\underline{\text{ }}\right)$}=\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}
\frac{3}{8}-9=-\frac{69}{8}: \\
6 \left(x^2-\frac{x}{2}+\frac{1}{16}\right)+9 \left(y^2-\frac{7 y}{9}+\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 }9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-7}{9}}{2}\right)^2=\frac{49}{324} \text{on }\text{the }\text{left }\text{and }9\times \frac{49}{324}=\frac{49}{36} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{49}{36}-\frac{69}{8}=-\frac{523}{72}: \\
6 \left(x^2-\frac{x}{2}+\frac{1}{16}\right)+9 \left(y^2-\frac{7 y}{9}+\frac{49}{324}\right)=\fbox{$-\frac{523}{72}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{x}{2}+\frac{1}{16}=\left(x-\frac{1}{4}\right)^2: \\
6 \fbox{$\left(x-\frac{1}{4}\right)^2$}+9 \left(y^2-\frac{7 y}{9}+\frac{49}{324}\right)=-\frac{523}{72} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{7 y}{9}+\frac{49}{324}=\left(y-\frac{7}{18}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 6 \left(x-\frac{1}{4}\right)^2+9 \fbox{$\left(y-\frac{7}{18}\right)^2$}=-\frac{523}{72} \\
\end{array}
| khanacademy | amps |
Given the equation $-3 x^2+2 x+y^2-10 y-5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
y^2-10 y-3 x^2+2 x-5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }5 \text{to }\text{both }\text{sides}: \\
y^2-10 y-3 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(-3 x^2+2 x+\underline{\text{ }}\right)+\left(y^2-10 y+\underline{\text{ }}\right)=\underline{\text{ }}+5 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-3 x^2+2 x+\underline{\text{ }}\right)=-3 \left(x^2-\frac{2 x}{3}+\underline{\text{ }}\right): \\
\fbox{$-3 \left(x^2-\frac{2 x}{3}+\underline{\text{ }}\right)$}+\left(y^2-10 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 }-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 6:
\begin{array}{l}
5-\frac{1}{3}=\frac{14}{3}: \\
-3 \left(x^2-\frac{2 x}{3}+\frac{1}{9}\right)+\left(y^2-10 y+\underline{\text{ }}\right)=\fbox{$\frac{14}{3}$} \\
\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{-10}{2}\right)^2=25 \text{to }\text{both }\text{sides}: \\
\end{array}
Step 8:
\begin{array}{l}
\frac{14}{3}+25=\frac{89}{3}: \\
-3 \left(x^2-\frac{2 x}{3}+\frac{1}{9}\right)+\left(y^2-10 y+25\right)=\fbox{$\frac{89}{3}$} \\
\end{array}
Step 9:
\begin{array}{l}
x^2-\frac{2 x}{3}+\frac{1}{9}=\left(x-\frac{1}{3}\right)^2: \\
-3 \fbox{$\left(x-\frac{1}{3}\right)^2$}+\left(y^2-10 y+25\right)=\frac{89}{3} \\
\end{array}
Step 10:
\begin{array}{l}
y^2-10 y+25=(y-5)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -3 \left(x-\frac{1}{3}\right)^2+\fbox{$(y-5)^2$}=\frac{89}{3} \\
\end{array}
| khanacademy | amps |
Given the equation $-10 y^2+6 y-9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-10 y^2+6 y-9=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }-10 y^2+6 y-9 \text{from }\text{both }\text{sides}: \\
10 y^2-6 y+9=0 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Subtract }9 \text{from }\text{both }\text{sides}: \\
10 y^2-6 y=-9 \\
\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(10 y^2-6 y+\underline{\text{ }}\right)=\underline{\text{ }}-9 \\
\end{array}
Step 5:
\begin{array}{l}
\left(10 y^2-6 y+\underline{\text{ }}\right)=10 \left(y^2-\frac{3 y}{5}+\underline{\text{ }}\right): \\
\fbox{$10 \left(y^2-\frac{3 y}{5}+\underline{\text{ }}\right)$}=\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 }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{-3}{5}}{2}\right)^2=\frac{9}{100} \text{on }\text{the }\text{left }\text{and }10\times \frac{9}{100}=\frac{9}{10} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{9}{10}-9=-\frac{81}{10}: \\
10 \left(y^2-\frac{3 y}{5}+\frac{9}{100}\right)=\fbox{$-\frac{81}{10}$} \\
\end{array}
Step 8:
\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{} & \text{10 }\fbox{$\left(y-\frac{3}{10}\right)^2$}=-\frac{81}{10} \\
\end{array}
| khanacademy | amps |
Given the equation $x^2-6 x+3 y^2-3 y+1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
3 y^2-3 y+x^2-6 x+1=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }1 \text{from }\text{both }\text{sides}: \\
3 y^2-3 y+x^2-6 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(x^2-6 x+\underline{\text{ }}\right)+\left(3 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\end{array}
Step 4:
\begin{array}{l}
\left(3 y^2-3 y+\underline{\text{ }}\right)=3 \left(y^2-y+\underline{\text{ }}\right): \\
\left(x^2-6 x+\underline{\text{ }}\right)+\fbox{$3 \left(y^2-y+\underline{\text{ }}\right)$}=\underline{\text{ }}-1 \\
\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{-6}{2}\right)^2=9 \text{to }\text{both }\text{sides}: \\
\end{array}
Step 6:
\begin{array}{l}
9-1=8: \\
\left(x^2-6 x+9\right)+3 \left(y^2-y+\underline{\text{ }}\right)=\fbox{$8$} \\
\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{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 8:
\begin{array}{l}
8+\frac{3}{4}=\frac{35}{4}: \\
\left(x^2-6 x+9\right)+3 \left(y^2-y+\frac{1}{4}\right)=\fbox{$\frac{35}{4}$} \\
\end{array}
Step 9:
\begin{array}{l}
x^2-6 x+9=(x-3)^2: \\
\fbox{$(x-3)^2$}+3 \left(y^2-y+\frac{1}{4}\right)=\frac{35}{4} \\
\end{array}
Step 10:
\begin{array}{l}
y^2-y+\frac{1}{4}=\left(y-\frac{1}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & (x-3)^2+3 \fbox{$\left(y-\frac{1}{2}\right)^2$}=\frac{35}{4} \\
\end{array}
| khanacademy | amps |
Given the equation $-7 x^2+3 x+5 y^2-2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-7 x^2+3 x+\left(5 y^2-2\right)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }5 y^2-7 x^2+3 x-2 \text{from }\text{both }\text{sides}: \\
7 x^2-3 x+\left(2-5 y^2\right)=0 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Subtract }2 \text{from }\text{both }\text{sides}: \\
-5 y^2+7 x^2-3 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(7 x^2-3 x+\underline{\text{ }}\right)-5 y^2=\underline{\text{ }}-2 \\
\end{array}
Step 5:
\begin{array}{l}
\left(7 x^2-3 x+\underline{\text{ }}\right)=7 \left(x^2-\frac{3 x}{7}+\underline{\text{ }}\right): \\
\fbox{$7 \left(x^2-\frac{3 x}{7}+\underline{\text{ }}\right)$}-5 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 }7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-3}{7}}{2}\right)^2=\frac{9}{196} \text{on }\text{the }\text{left }\text{and }7\times \frac{9}{196}=\frac{9}{28} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{9}{28}-2=-\frac{47}{28}: \\
7 \left(x^2-\frac{3 x}{7}+\frac{9}{196}\right)-5 y^2=\fbox{$-\frac{47}{28}$} \\
\end{array}
Step 8:
\begin{array}{l}
x^2-\frac{3 x}{7}+\frac{9}{196}=\left(x-\frac{3}{14}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 7 \fbox{$\left(x-\frac{3}{14}\right)^2$}-5 y^2=-\frac{47}{28} \\
\end{array}
| khanacademy | amps |
Given the equation $8 x^2+6 x+8 y^2+10 y+8=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
8 y^2+10 y+8 x^2+6 x+8=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }8 \text{from }\text{both }\text{sides}: \\
8 y^2+10 y+8 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(8 x^2+6 x+\underline{\text{ }}\right)+\left(8 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}-8 \\
\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(8 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}-8 \\
\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): \\
8 \left(x^2+\frac{3 x}{4}+\underline{\text{ }}\right)+\fbox{$8 \left(y^2+\frac{5 y}{4}+\underline{\text{ }}\right)$}=\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 }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}
\frac{9}{8}-8=-\frac{55}{8}: \\
8 \left(x^2+\frac{3 x}{4}+\frac{9}{64}\right)+8 \left(y^2+\frac{5 y}{4}+\underline{\text{ }}\right)=\fbox{$-\frac{55}{8}$} \\
\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 }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{55}{8}=-\frac{15}{4}: \\
8 \left(x^2+\frac{3 x}{4}+\frac{9}{64}\right)+8 \left(y^2+\frac{5 y}{4}+\frac{25}{64}\right)=\fbox{$-\frac{15}{4}$} \\
\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{$\left(x+\frac{3}{8}\right)^2$}+8 \left(y^2+\frac{5 y}{4}+\frac{25}{64}\right)=-\frac{15}{4} \\
\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{} & 8 \left(x+\frac{3}{8}\right)^2+8 \fbox{$\left(y+\frac{5}{8}\right)^2$}=-\frac{15}{4} \\
\end{array}
| khanacademy | amps |
Given the equation $2 x^2+8 x-5 y^2+7 y+3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-5 y^2+7 y+2 x^2+8 x+3=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }3 \text{from }\text{both }\text{sides}: \\
-5 y^2+7 y+2 x^2+8 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(2 x^2+8 x+\underline{\text{ }}\right)+\left(-5 y^2+7 y+\underline{\text{ }}\right)=\underline{\text{ }}-3 \\
\end{array}
Step 4:
\begin{array}{l}
\left(2 x^2+8 x+\underline{\text{ }}\right)=2 \left(x^2+4 x+\underline{\text{ }}\right): \\
\fbox{$2 \left(x^2+4 x+\underline{\text{ }}\right)$}+\left(-5 y^2+7 y+\underline{\text{ }}\right)=\underline{\text{ }}-3 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-5 y^2+7 y+\underline{\text{ }}\right)=-5 \left(y^2-\frac{7 y}{5}+\underline{\text{ }}\right): \\
2 \left(x^2+4 x+\underline{\text{ }}\right)+\fbox{$-5 \left(y^2-\frac{7 y}{5}+\underline{\text{ }}\right)$}=\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 }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 7:
\begin{array}{l}
8-3=5: \\
2 \left(x^2+4 x+4\right)-5 \left(y^2-\frac{7 y}{5}+\underline{\text{ }}\right)=\fbox{$5$} \\
\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 }-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 9:
\begin{array}{l}
5-\frac{49}{20}=\frac{51}{20}: \\
2 \left(x^2+4 x+4\right)-5 \left(y^2-\frac{7 y}{5}+\frac{49}{100}\right)=\fbox{$\frac{51}{20}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+4 x+4=(x+2)^2: \\
2 \fbox{$(x+2)^2$}-5 \left(y^2-\frac{7 y}{5}+\frac{49}{100}\right)=\frac{51}{20} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{7 y}{5}+\frac{49}{100}=\left(y-\frac{7}{10}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 2 (x+2)^2-5 \fbox{$\left(y-\frac{7}{10}\right)^2$}=\frac{51}{20} \\
\end{array}
| khanacademy | amps |
Given the equation $4 x^2+2 x-y^2-5 y+2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-y^2-5 y+4 x^2+2 x+2=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }2 \text{from }\text{both }\text{sides}: \\
-y^2-5 y+4 x^2+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(4 x^2+2 x+\underline{\text{ }}\right)+\left(-y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}-2 \\
\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(-y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}-2 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-y^2-5 y+\underline{\text{ }}\right)=-\left(y^2+5 y+\underline{\text{ }}\right): \\
4 \left(x^2+\frac{x}{2}+\underline{\text{ }}\right)+\fbox{$-\left(y^2+5 y+\underline{\text{ }}\right)$}=\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 }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}
\frac{1}{4}-2=-\frac{7}{4}: \\
4 \left(x^2+\frac{x}{2}+\frac{1}{16}\right)-\left(y^2+5 y+\underline{\text{ }}\right)=\fbox{$-\frac{7}{4}$} \\
\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 }-1 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{5}{2}\right)^2=\frac{25}{4} \text{on }\text{the }\text{left }\text{and }-\frac{25}{4}=-\frac{25}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{7}{4}-\frac{25}{4}=-8: \\
4 \left(x^2+\frac{x}{2}+\frac{1}{16}\right)-\left(y^2+5 y+\frac{25}{4}\right)=\fbox{$-8$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{x}{2}+\frac{1}{16}=\left(x+\frac{1}{4}\right)^2: \\
4 \fbox{$\left(x+\frac{1}{4}\right)^2$}-\left(y^2+5 y+\frac{25}{4}\right)=-8 \\
\end{array}
Step 11:
\begin{array}{l}
y^2+5 y+\frac{25}{4}=\left(y+\frac{5}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 4 \left(x+\frac{1}{4}\right)^2-\fbox{$\left(y+\frac{5}{2}\right)^2$}=-8 \\
\end{array}
| khanacademy | amps |
Given the equation $-9 x^2+7 x+8 y^2-3 y-2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
8 y^2-3 y-9 x^2+7 x-2=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }2 \text{to }\text{both }\text{sides}: \\
8 y^2-3 y-9 x^2+7 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(-9 x^2+7 x+\underline{\text{ }}\right)+\left(8 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}+2 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-9 x^2+7 x+\underline{\text{ }}\right)=-9 \left(x^2-\frac{7 x}{9}+\underline{\text{ }}\right): \\
\fbox{$-9 \left(x^2-\frac{7 x}{9}+\underline{\text{ }}\right)$}+\left(8 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}+2 \\
\end{array}
Step 5:
\begin{array}{l}
\left(8 y^2-3 y+\underline{\text{ }}\right)=8 \left(y^2-\frac{3 y}{8}+\underline{\text{ }}\right): \\
-9 \left(x^2-\frac{7 x}{9}+\underline{\text{ }}\right)+\fbox{$8 \left(y^2-\frac{3 y}{8}+\underline{\text{ }}\right)$}=\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 }-9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-7}{9}}{2}\right)^2=\frac{49}{324} \text{on }\text{the }\text{left }\text{and }-9\times \frac{49}{324}=-\frac{49}{36} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
2-\frac{49}{36}=\frac{23}{36}: \\
-9 \left(x^2-\frac{7 x}{9}+\frac{49}{324}\right)+8 \left(y^2-\frac{3 y}{8}+\underline{\text{ }}\right)=\fbox{$\frac{23}{36}$} \\
\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{-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 9:
\begin{array}{l}
\frac{23}{36}+\frac{9}{32}=\frac{265}{288}: \\
-9 \left(x^2-\frac{7 x}{9}+\frac{49}{324}\right)+8 \left(y^2-\frac{3 y}{8}+\frac{9}{256}\right)=\fbox{$\frac{265}{288}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{7 x}{9}+\frac{49}{324}=\left(x-\frac{7}{18}\right)^2: \\
-9 \fbox{$\left(x-\frac{7}{18}\right)^2$}+8 \left(y^2-\frac{3 y}{8}+\frac{9}{256}\right)=\frac{265}{288} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{3 y}{8}+\frac{9}{256}=\left(y-\frac{3}{16}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -9 \left(x-\frac{7}{18}\right)^2+8 \fbox{$\left(y-\frac{3}{16}\right)^2$}=\frac{265}{288} \\
\end{array}
| khanacademy | amps |
Given the equation $9 x^2+5 x+3 y^2+8 y-5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
3 y^2+8 y+9 x^2+5 x-5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }5 \text{to }\text{both }\text{sides}: \\
3 y^2+8 y+9 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(9 x^2+5 x+\underline{\text{ }}\right)+\left(3 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}+5 \\
\end{array}
Step 4:
\begin{array}{l}
\left(9 x^2+5 x+\underline{\text{ }}\right)=9 \left(x^2+\frac{5 x}{9}+\underline{\text{ }}\right): \\
\fbox{$9 \left(x^2+\frac{5 x}{9}+\underline{\text{ }}\right)$}+\left(3 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}+5 \\
\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): \\
9 \left(x^2+\frac{5 x}{9}+\underline{\text{ }}\right)+\fbox{$3 \left(y^2+\frac{8 y}{3}+\underline{\text{ }}\right)$}=\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 }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 7:
\begin{array}{l}
5+\frac{25}{36}=\frac{205}{36}: \\
9 \left(x^2+\frac{5 x}{9}+\frac{25}{324}\right)+3 \left(y^2+\frac{8 y}{3}+\underline{\text{ }}\right)=\fbox{$\frac{205}{36}$} \\
\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{205}{36}+\frac{16}{3}=\frac{397}{36}: \\
9 \left(x^2+\frac{5 x}{9}+\frac{25}{324}\right)+3 \left(y^2+\frac{8 y}{3}+\frac{16}{9}\right)=\fbox{$\frac{397}{36}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{5 x}{9}+\frac{25}{324}=\left(x+\frac{5}{18}\right)^2: \\
9 \fbox{$\left(x+\frac{5}{18}\right)^2$}+3 \left(y^2+\frac{8 y}{3}+\frac{16}{9}\right)=\frac{397}{36} \\
\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{} & 9 \left(x+\frac{5}{18}\right)^2+3 \fbox{$\left(y+\frac{4}{3}\right)^2$}=\frac{397}{36} \\
\end{array}
| khanacademy | amps |
Given the equation $-2 x^2+4 x-8 y-8=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-2 x^2+4 x+(-8 y-8)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }-8 y-2 x^2+4 x-8 \text{from }\text{both }\text{sides}: \\
2 x^2-4 x+(8 y+8)=0 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Subtract }8 y+8 \text{from }\text{both }\text{sides}: \\
2 x^2-4 x=-8 y-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(2 x^2-4 x+\underline{\text{ }}\right)=(-8 y-8)+\underline{\text{ }} \\
\end{array}
Step 5:
\begin{array}{l}
\left(2 x^2-4 x+\underline{\text{ }}\right)=2 \left(x^2-2 x+\underline{\text{ }}\right): \\
\fbox{$2 \left(x^2-2 x+\underline{\text{ }}\right)$}=(-8 y-8)+\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{-2}{2}\right)^2=1 \text{on }\text{the }\text{left }\text{and }2\times 1=2 \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
(-8 y-8)+2=-8 y-6: \\
2 \left(x^2-2 x+1\right)=\fbox{$-8 y-6$} \\
\end{array}
Step 8:
\begin{array}{l}
x^2-2 x+1=(x-1)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 2 \fbox{$(x-1)^2$}=-8 y-6 \\
\end{array}
| khanacademy | amps |
Given the equation $-3 x^2+4 x+6 y^2+8 y-5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
6 y^2+8 y-3 x^2+4 x-5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }5 \text{to }\text{both }\text{sides}: \\
6 y^2+8 y-3 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(-3 x^2+4 x+\underline{\text{ }}\right)+\left(6 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}+5 \\
\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(6 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}+5 \\
\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): \\
-3 \left(x^2-\frac{4 x}{3}+\underline{\text{ }}\right)+\fbox{$6 \left(y^2+\frac{4 y}{3}+\underline{\text{ }}\right)$}=\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 }-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}
5-\frac{4}{3}=\frac{11}{3}: \\
-3 \left(x^2-\frac{4 x}{3}+\frac{4}{9}\right)+6 \left(y^2+\frac{4 y}{3}+\underline{\text{ }}\right)=\fbox{$\frac{11}{3}$} \\
\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 }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{11}{3}+\frac{8}{3}=\frac{19}{3}: \\
-3 \left(x^2-\frac{4 x}{3}+\frac{4}{9}\right)+6 \left(y^2+\frac{4 y}{3}+\frac{4}{9}\right)=\fbox{$\frac{19}{3}$} \\
\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{$\left(x-\frac{2}{3}\right)^2$}+6 \left(y^2+\frac{4 y}{3}+\frac{4}{9}\right)=\frac{19}{3} \\
\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{} & -3 \left(x-\frac{2}{3}\right)^2+6 \fbox{$\left(y+\frac{2}{3}\right)^2$}=\frac{19}{3} \\
\end{array}
| khanacademy | amps |
Given the equation $-3 x^2-3 x+2 y^2+5 y+7=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
2 y^2+5 y-3 x^2-3 x+7=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }7 \text{from }\text{both }\text{sides}: \\
2 y^2+5 y-3 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(-3 x^2-3 x+\underline{\text{ }}\right)+\left(2 y^2+5 y+\underline{\text{ }}\right)=\underline{\text{ }}-7 \\
\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(2 y^2+5 y+\underline{\text{ }}\right)=\underline{\text{ }}-7 \\
\end{array}
Step 5:
\begin{array}{l}
\left(2 y^2+5 y+\underline{\text{ }}\right)=2 \left(y^2+\frac{5 y}{2}+\underline{\text{ }}\right): \\
-3 \left(x^2+x+\underline{\text{ }}\right)+\fbox{$2 \left(y^2+\frac{5 y}{2}+\underline{\text{ }}\right)$}=\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 }-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}
-7-\frac{3}{4}=-\frac{31}{4}: \\
-3 \left(x^2+x+\frac{1}{4}\right)+2 \left(y^2+\frac{5 y}{2}+\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 }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 9:
\begin{array}{l}
\frac{25}{8}-\frac{31}{4}=-\frac{37}{8}: \\
-3 \left(x^2+x+\frac{1}{4}\right)+2 \left(y^2+\frac{5 y}{2}+\frac{25}{16}\right)=\fbox{$-\frac{37}{8}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+x+\frac{1}{4}=\left(x+\frac{1}{2}\right)^2: \\
-3 \fbox{$\left(x+\frac{1}{2}\right)^2$}+2 \left(y^2+\frac{5 y}{2}+\frac{25}{16}\right)=-\frac{37}{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{} & -3 \left(x+\frac{1}{2}\right)^2+2 \fbox{$\left(y+\frac{5}{4}\right)^2$}=-\frac{37}{8} \\
\end{array}
| khanacademy | amps |
Given the equation $-x^2-x+2 y^2-6 y-10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
2 y^2-6 y-x^2-x-10=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }\text{10 }\text{to }\text{both }\text{sides}: \\
2 y^2-6 y-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(-x^2-x+\underline{\text{ }}\right)+\left(2 y^2-6 y+\underline{\text{ }}\right)=\underline{\text{ }}+10 \\
\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(2 y^2-6 y+\underline{\text{ }}\right)=\underline{\text{ }}+10 \\
\end{array}
Step 5:
\begin{array}{l}
\left(2 y^2-6 y+\underline{\text{ }}\right)=2 \left(y^2-3 y+\underline{\text{ }}\right): \\
-\left(x^2+x+\underline{\text{ }}\right)+\fbox{$2 \left(y^2-3 y+\underline{\text{ }}\right)$}=\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{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}
10-\frac{1}{4}=\frac{39}{4}: \\
-\left(x^2+x+\frac{1}{4}\right)+2 \left(y^2-3 y+\underline{\text{ }}\right)=\fbox{$\frac{39}{4}$} \\
\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 }2 \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 }2\times \frac{9}{4}=\frac{9}{2} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{39}{4}+\frac{9}{2}=\frac{57}{4}: \\
-\left(x^2+x+\frac{1}{4}\right)+2 \left(y^2-3 y+\frac{9}{4}\right)=\fbox{$\frac{57}{4}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+x+\frac{1}{4}=\left(x+\frac{1}{2}\right)^2: \\
-\fbox{$\left(x+\frac{1}{2}\right)^2$}+2 \left(y^2-3 y+\frac{9}{4}\right)=\frac{57}{4} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-3 y+\frac{9}{4}=\left(y-\frac{3}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -\left(x+\frac{1}{2}\right)^2+2 \fbox{$\left(y-\frac{3}{2}\right)^2$}=\frac{57}{4} \\
\end{array}
| khanacademy | amps |
Given the equation $5 x^2-3 x+9 y^2+7 y=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
9 y^2+7 y+5 x^2-3 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(5 x^2-3 x+\underline{\text{ }}\right)+\left(9 y^2+7 y+\underline{\text{ }}\right)=\underline{\text{ }}+0 \\
\end{array}
Step 3:
\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(9 y^2+7 y+\underline{\text{ }}\right)=\underline{\text{ }}+0 \\
\end{array}
Step 4:
\begin{array}{l}
\left(9 y^2+7 y+\underline{\text{ }}\right)=9 \left(y^2+\frac{7 y}{9}+\underline{\text{ }}\right): \\
5 \left(x^2-\frac{3 x}{5}+\underline{\text{ }}\right)+\fbox{$9 \left(y^2+\frac{7 y}{9}+\underline{\text{ }}\right)$}=\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 }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 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{7}{9}}{2}\right)^2=\frac{49}{324} \text{on }\text{the }\text{left }\text{and }9\times \frac{49}{324}=\frac{49}{36} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{9}{20}+\frac{49}{36}=\frac{163}{90}: \\
5 \left(x^2-\frac{3 x}{5}+\frac{9}{100}\right)+9 \left(y^2+\frac{7 y}{9}+\frac{49}{324}\right)=\fbox{$\frac{163}{90}$} \\
\end{array}
Step 8:
\begin{array}{l}
x^2-\frac{3 x}{5}+\frac{9}{100}=\left(x-\frac{3}{10}\right)^2: \\
5 \fbox{$\left(x-\frac{3}{10}\right)^2$}+9 \left(y^2+\frac{7 y}{9}+\frac{49}{324}\right)=\frac{163}{90} \\
\end{array}
Step 9:
\begin{array}{l}
y^2+\frac{7 y}{9}+\frac{49}{324}=\left(y+\frac{7}{18}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 5 \left(x-\frac{3}{10}\right)^2+9 \fbox{$\left(y+\frac{7}{18}\right)^2$}=\frac{163}{90} \\
\end{array}
| khanacademy | amps |
Given the equation $4 x^2-9 x+3 y^2+4=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
4 x^2-9 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+4 x^2-9 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(4 x^2-9 x+\underline{\text{ }}\right)+3 y^2=\underline{\text{ }}-4 \\
\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{$4 \left(x^2-\frac{9 x}{4}+\underline{\text{ }}\right)$}+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 }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}
\frac{81}{16}-4=\frac{17}{16}: \\
4 \left(x^2-\frac{9 x}{4}+\frac{81}{64}\right)+3 y^2=\fbox{$\frac{17}{16}$} \\
\end{array}
Step 7:
\begin{array}{l}
x^2-\frac{9 x}{4}+\frac{81}{64}=\left(x-\frac{9}{8}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 4 \fbox{$\left(x-\frac{9}{8}\right)^2$}+3 y^2=\frac{17}{16} \\
\end{array}
| khanacademy | amps |
Given the equation $-2 x^2-8 x+7 y^2+5 y-1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
7 y^2+5 y-2 x^2-8 x-1=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }1 \text{to }\text{both }\text{sides}: \\
7 y^2+5 y-2 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(-2 x^2-8 x+\underline{\text{ }}\right)+\left(7 y^2+5 y+\underline{\text{ }}\right)=\underline{\text{ }}+1 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-2 x^2-8 x+\underline{\text{ }}\right)=-2 \left(x^2+4 x+\underline{\text{ }}\right): \\
\fbox{$-2 \left(x^2+4 x+\underline{\text{ }}\right)$}+\left(7 y^2+5 y+\underline{\text{ }}\right)=\underline{\text{ }}+1 \\
\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+4 x+\underline{\text{ }}\right)+\fbox{$7 \left(y^2+\frac{5 y}{7}+\underline{\text{ }}\right)$}=\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 }-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 7:
\begin{array}{l}
1-8=-7: \\
-2 \left(x^2+4 x+4\right)+7 \left(y^2+\frac{5 y}{7}+\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 }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}-7=-\frac{171}{28}: \\
-2 \left(x^2+4 x+4\right)+7 \left(y^2+\frac{5 y}{7}+\frac{25}{196}\right)=\fbox{$-\frac{171}{28}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+4 x+4=(x+2)^2: \\
-2 \fbox{$(x+2)^2$}+7 \left(y^2+\frac{5 y}{7}+\frac{25}{196}\right)=-\frac{171}{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 (x+2)^2+7 \fbox{$\left(y+\frac{5}{14}\right)^2$}=-\frac{171}{28} \\
\end{array}
| khanacademy | amps |
Given the equation $10 x^2+2 y^2-2 y+6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
2 y^2-2 y+\left(10 x^2+6\right)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }6 \text{from }\text{both }\text{sides}: \\
2 y^2-2 y+10 x^2=-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(2 y^2-2 y+\underline{\text{ }}\right)+10 x^2=\underline{\text{ }}-6 \\
\end{array}
Step 4:
\begin{array}{l}
\left(2 y^2-2 y+\underline{\text{ }}\right)=2 \left(y^2-y+\underline{\text{ }}\right): \\
\fbox{$2 \left(y^2-y+\underline{\text{ }}\right)$}+10 x^2=\underline{\text{ }}-6 \\
\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 }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 6:
\begin{array}{l}
\frac{1}{2}-6=-\frac{11}{2}: \\
2 \left(y^2-y+\frac{1}{4}\right)+10 x^2=\fbox{$-\frac{11}{2}$} \\
\end{array}
Step 7:
\begin{array}{l}
y^2-y+\frac{1}{4}=\left(y-\frac{1}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 2 \fbox{$\left(y-\frac{1}{2}\right)^2$}+10 x^2=-\frac{11}{2} \\
\end{array}
| khanacademy | amps |
Given the equation $7 x^2+2 x+7 y^2+8 y-6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
7 y^2+8 y+7 x^2+2 x-6=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }6 \text{to }\text{both }\text{sides}: \\
7 y^2+8 y+7 x^2+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(7 x^2+2 x+\underline{\text{ }}\right)+\left(7 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\end{array}
Step 4:
\begin{array}{l}
\left(7 x^2+2 x+\underline{\text{ }}\right)=7 \left(x^2+\frac{2 x}{7}+\underline{\text{ }}\right): \\
\fbox{$7 \left(x^2+\frac{2 x}{7}+\underline{\text{ }}\right)$}+\left(7 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\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): \\
7 \left(x^2+\frac{2 x}{7}+\underline{\text{ }}\right)+\fbox{$7 \left(y^2+\frac{8 y}{7}+\underline{\text{ }}\right)$}=\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{2}{7}}{2}\right)^2=\frac{1}{49} \text{on }\text{the }\text{left }\text{and }\frac{7}{49}=\frac{1}{7} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
6+\frac{1}{7}=\frac{43}{7}: \\
7 \left(x^2+\frac{2 x}{7}+\frac{1}{49}\right)+7 \left(y^2+\frac{8 y}{7}+\underline{\text{ }}\right)=\fbox{$\frac{43}{7}$} \\
\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{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{43}{7}+\frac{16}{7}=\frac{59}{7}: \\
7 \left(x^2+\frac{2 x}{7}+\frac{1}{49}\right)+7 \left(y^2+\frac{8 y}{7}+\frac{16}{49}\right)=\fbox{$\frac{59}{7}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{2 x}{7}+\frac{1}{49}=\left(x+\frac{1}{7}\right)^2: \\
7 \fbox{$\left(x+\frac{1}{7}\right)^2$}+7 \left(y^2+\frac{8 y}{7}+\frac{16}{49}\right)=\frac{59}{7} \\
\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{} & 7 \left(x+\frac{1}{7}\right)^2+7 \fbox{$\left(y+\frac{4}{7}\right)^2$}=\frac{59}{7} \\
\end{array}
| khanacademy | amps |
Given the equation $3 x^2+7 x-10 y^2-4 y-2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-10 y^2-4 y+3 x^2+7 x-2=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }2 \text{to }\text{both }\text{sides}: \\
-10 y^2-4 y+3 x^2+7 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+7 x+\underline{\text{ }}\right)+\left(-10 y^2-4 y+\underline{\text{ }}\right)=\underline{\text{ }}+2 \\
\end{array}
Step 4:
\begin{array}{l}
\left(3 x^2+7 x+\underline{\text{ }}\right)=3 \left(x^2+\frac{7 x}{3}+\underline{\text{ }}\right): \\
\fbox{$3 \left(x^2+\frac{7 x}{3}+\underline{\text{ }}\right)$}+\left(-10 y^2-4 y+\underline{\text{ }}\right)=\underline{\text{ }}+2 \\
\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): \\
3 \left(x^2+\frac{7 x}{3}+\underline{\text{ }}\right)+\fbox{$-10 \left(y^2+\frac{2 y}{5}+\underline{\text{ }}\right)$}=\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 }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 7:
\begin{array}{l}
2+\frac{49}{12}=\frac{73}{12}: \\
3 \left(x^2+\frac{7 x}{3}+\frac{49}{36}\right)-10 \left(y^2+\frac{2 y}{5}+\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 }-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{73}{12}-\frac{2}{5}=\frac{341}{60}: \\
3 \left(x^2+\frac{7 x}{3}+\frac{49}{36}\right)-10 \left(y^2+\frac{2 y}{5}+\frac{1}{25}\right)=\fbox{$\frac{341}{60}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{7 x}{3}+\frac{49}{36}=\left(x+\frac{7}{6}\right)^2: \\
3 \fbox{$\left(x+\frac{7}{6}\right)^2$}-10 \left(y^2+\frac{2 y}{5}+\frac{1}{25}\right)=\frac{341}{60} \\
\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{} & 3 \left(x+\frac{7}{6}\right)^2-\text{10 }\fbox{$\left(y+\frac{1}{5}\right)^2$}=\frac{341}{60} \\
\end{array}
| khanacademy | amps |
Given the equation $9 x^2-3 x-2 y^2+y=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-2 y^2+y+9 x^2-3 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(9 x^2-3 x+\underline{\text{ }}\right)+\left(-2 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}+0 \\
\end{array}
Step 3:
\begin{array}{l}
\left(9 x^2-3 x+\underline{\text{ }}\right)=9 \left(x^2-\frac{x}{3}+\underline{\text{ }}\right): \\
\fbox{$9 \left(x^2-\frac{x}{3}+\underline{\text{ }}\right)$}+\left(-2 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}+0 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-2 y^2+y+\underline{\text{ }}\right)=-2 \left(y^2-\frac{y}{2}+\underline{\text{ }}\right): \\
9 \left(x^2-\frac{x}{3}+\underline{\text{ }}\right)+\fbox{$-2 \left(y^2-\frac{y}{2}+\underline{\text{ }}\right)$}=\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 }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 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 }-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}
\frac{1}{4}-\frac{1}{8}=\frac{1}{8}: \\
9 \left(x^2-\frac{x}{3}+\frac{1}{36}\right)-2 \left(y^2-\frac{y}{2}+\frac{1}{16}\right)=\fbox{$\frac{1}{8}$} \\
\end{array}
Step 8:
\begin{array}{l}
x^2-\frac{x}{3}+\frac{1}{36}=\left(x-\frac{1}{6}\right)^2: \\
9 \fbox{$\left(x-\frac{1}{6}\right)^2$}-2 \left(y^2-\frac{y}{2}+\frac{1}{16}\right)=\frac{1}{8} \\
\end{array}
Step 9:
\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{} & 9 \left(x-\frac{1}{6}\right)^2-2 \fbox{$\left(y-\frac{1}{4}\right)^2$}=\frac{1}{8} \\
\end{array}
| khanacademy | amps |
Given the equation $8 x^2-7 x+8 y^2+7 y+5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
8 y^2+7 y+8 x^2-7 x+5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }5 \text{from }\text{both }\text{sides}: \\
8 y^2+7 y+8 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(8 x^2-7 x+\underline{\text{ }}\right)+\left(8 y^2+7 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\end{array}
Step 4:
\begin{array}{l}
\left(8 x^2-7 x+\underline{\text{ }}\right)=8 \left(x^2-\frac{7 x}{8}+\underline{\text{ }}\right): \\
\fbox{$8 \left(x^2-\frac{7 x}{8}+\underline{\text{ }}\right)$}+\left(8 y^2+7 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\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): \\
8 \left(x^2-\frac{7 x}{8}+\underline{\text{ }}\right)+\fbox{$8 \left(y^2+\frac{7 y}{8}+\underline{\text{ }}\right)$}=\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 }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 7:
\begin{array}{l}
\frac{49}{32}-5=-\frac{111}{32}: \\
8 \left(x^2-\frac{7 x}{8}+\frac{49}{256}\right)+8 \left(y^2+\frac{7 y}{8}+\underline{\text{ }}\right)=\fbox{$-\frac{111}{32}$} \\
\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 }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{49}{32}-\frac{111}{32}=-\frac{31}{16}: \\
8 \left(x^2-\frac{7 x}{8}+\frac{49}{256}\right)+8 \left(y^2+\frac{7 y}{8}+\frac{49}{256}\right)=\fbox{$-\frac{31}{16}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{7 x}{8}+\frac{49}{256}=\left(x-\frac{7}{16}\right)^2: \\
8 \fbox{$\left(x-\frac{7}{16}\right)^2$}+8 \left(y^2+\frac{7 y}{8}+\frac{49}{256}\right)=-\frac{31}{16} \\
\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{} & 8 \left(x-\frac{7}{16}\right)^2+8 \fbox{$\left(y+\frac{7}{16}\right)^2$}=-\frac{31}{16} \\
\end{array}
| khanacademy | amps |
Given the equation $-x^2-x+3 y^2+6 y+5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
3 y^2+6 y-x^2-x+5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }5 \text{from }\text{both }\text{sides}: \\
3 y^2+6 y-x^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(-x^2-x+\underline{\text{ }}\right)+\left(3 y^2+6 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\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(3 y^2+6 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\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): \\
-\left(x^2+x+\underline{\text{ }}\right)+\fbox{$3 \left(y^2+2 y+\underline{\text{ }}\right)$}=\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 }-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}
-5-\frac{1}{4}=-\frac{21}{4}: \\
-\left(x^2+x+\frac{1}{4}\right)+3 \left(y^2+2 y+\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 }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}
3-\frac{21}{4}=-\frac{9}{4}: \\
-\left(x^2+x+\frac{1}{4}\right)+3 \left(y^2+2 y+1\right)=\fbox{$-\frac{9}{4}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+x+\frac{1}{4}=\left(x+\frac{1}{2}\right)^2: \\
-\fbox{$\left(x+\frac{1}{2}\right)^2$}+3 \left(y^2+2 y+1\right)=-\frac{9}{4} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+2 y+1=(y+1)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -\left(x+\frac{1}{2}\right)^2+3 \fbox{$(y+1)^2$}=-\frac{9}{4} \\
\end{array}
| khanacademy | amps |
Given the equation $4 x^2+7 x+y^2+6 y-3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
y^2+6 y+4 x^2+7 x-3=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }3 \text{to }\text{both }\text{sides}: \\
y^2+6 y+4 x^2+7 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(4 x^2+7 x+\underline{\text{ }}\right)+\left(y^2+6 y+\underline{\text{ }}\right)=\underline{\text{ }}+3 \\
\end{array}
Step 4:
\begin{array}{l}
\left(4 x^2+7 x+\underline{\text{ }}\right)=4 \left(x^2+\frac{7 x}{4}+\underline{\text{ }}\right): \\
\fbox{$4 \left(x^2+\frac{7 x}{4}+\underline{\text{ }}\right)$}+\left(y^2+6 y+\underline{\text{ }}\right)=\underline{\text{ }}+3 \\
\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{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 6:
\begin{array}{l}
3+\frac{49}{16}=\frac{97}{16}: \\
4 \left(x^2+\frac{7 x}{4}+\frac{49}{64}\right)+\left(y^2+6 y+\underline{\text{ }}\right)=\fbox{$\frac{97}{16}$} \\
\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}
\frac{97}{16}+9=\frac{241}{16}: \\
4 \left(x^2+\frac{7 x}{4}+\frac{49}{64}\right)+\left(y^2+6 y+9\right)=\fbox{$\frac{241}{16}$} \\
\end{array}
Step 9:
\begin{array}{l}
x^2+\frac{7 x}{4}+\frac{49}{64}=\left(x+\frac{7}{8}\right)^2: \\
4 \fbox{$\left(x+\frac{7}{8}\right)^2$}+\left(y^2+6 y+9\right)=\frac{241}{16} \\
\end{array}
Step 10:
\begin{array}{l}
y^2+6 y+9=(y+3)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 4 \left(x+\frac{7}{8}\right)^2+\fbox{$(y+3)^2$}=\frac{241}{16} \\
\end{array}
| khanacademy | amps |
Given the equation $-6 x^2+x-4 y^2+10 y+4=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-4 y^2+10 y-6 x^2+x+4=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }4 \text{from }\text{both }\text{sides}: \\
-4 y^2+10 y-6 x^2+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(-6 x^2+x+\underline{\text{ }}\right)+\left(-4 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}-4 \\
\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(-4 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}-4 \\
\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): \\
-6 \left(x^2-\frac{x}{6}+\underline{\text{ }}\right)+\fbox{$-4 \left(y^2-\frac{5 y}{2}+\underline{\text{ }}\right)$}=\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 }-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}
-4-\frac{1}{24}=-\frac{97}{24}: \\
-6 \left(x^2-\frac{x}{6}+\frac{1}{144}\right)-4 \left(y^2-\frac{5 y}{2}+\underline{\text{ }}\right)=\fbox{$-\frac{97}{24}$} \\
\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 }-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{97}{24}-\frac{25}{4}=-\frac{247}{24}: \\
-6 \left(x^2-\frac{x}{6}+\frac{1}{144}\right)-4 \left(y^2-\frac{5 y}{2}+\frac{25}{16}\right)=\fbox{$-\frac{247}{24}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{x}{6}+\frac{1}{144}=\left(x-\frac{1}{12}\right)^2: \\
-6 \fbox{$\left(x-\frac{1}{12}\right)^2$}-4 \left(y^2-\frac{5 y}{2}+\frac{25}{16}\right)=-\frac{247}{24} \\
\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{} & -6 \left(x-\frac{1}{12}\right)^2-4 \fbox{$\left(y-\frac{5}{4}\right)^2$}=-\frac{247}{24} \\
\end{array}
| khanacademy | amps |
Given the equation $-2 x^2-7 x-4 y^2-4 y+10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-4 y^2-4 y-2 x^2-7 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-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(-4 y^2-4 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(-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): \\
-2 \left(x^2+\frac{7 x}{2}+\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 }-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)-4 \left(y^2+y+\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 }-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{129}{8}-1=-\frac{137}{8}: \\
-2 \left(x^2+\frac{7 x}{2}+\frac{49}{16}\right)-4 \left(y^2+y+\frac{1}{4}\right)=\fbox{$-\frac{137}{8}$} \\
\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{$\left(x+\frac{7}{4}\right)^2$}-4 \left(y^2+y+\frac{1}{4}\right)=-\frac{137}{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{} & -2 \left(x+\frac{7}{4}\right)^2-4 \fbox{$\left(y+\frac{1}{2}\right)^2$}=-\frac{137}{8} \\
\end{array}
| khanacademy | amps |
Given the equation $10 x^2-x+2 y^2-7 y-6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
2 y^2-7 y+10 x^2-x-6=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }6 \text{to }\text{both }\text{sides}: \\
2 y^2-7 y+10 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(10 x^2-x+\underline{\text{ }}\right)+\left(2 y^2-7 y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\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(2 y^2-7 y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\end{array}
Step 5:
\begin{array}{l}
\left(2 y^2-7 y+\underline{\text{ }}\right)=2 \left(y^2-\frac{7 y}{2}+\underline{\text{ }}\right): \\
10 \left(x^2-\frac{x}{10}+\underline{\text{ }}\right)+\fbox{$2 \left(y^2-\frac{7 y}{2}+\underline{\text{ }}\right)$}=\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 }\text{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}
6+\frac{1}{40}=\frac{241}{40}: \\
10 \left(x^2-\frac{x}{10}+\frac{1}{400}\right)+2 \left(y^2-\frac{7 y}{2}+\underline{\text{ }}\right)=\fbox{$\frac{241}{40}$} \\
\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 }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 9:
\begin{array}{l}
\frac{241}{40}+\frac{49}{8}=\frac{243}{20}: \\
10 \left(x^2-\frac{x}{10}+\frac{1}{400}\right)+2 \left(y^2-\frac{7 y}{2}+\frac{49}{16}\right)=\fbox{$\frac{243}{20}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{x}{10}+\frac{1}{400}=\left(x-\frac{1}{20}\right)^2: \\
\text{10 }\fbox{$\left(x-\frac{1}{20}\right)^2$}+2 \left(y^2-\frac{7 y}{2}+\frac{49}{16}\right)=\frac{243}{20} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{7 y}{2}+\frac{49}{16}=\left(y-\frac{7}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 10 \left(x-\frac{1}{20}\right)^2+2 \fbox{$\left(y-\frac{7}{4}\right)^2$}=\frac{243}{20} \\
\end{array}
| khanacademy | amps |
Given the equation $-3 x^2+10 x-4 y^2+10 y+2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-4 y^2+10 y-3 x^2+10 x+2=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }2 \text{from }\text{both }\text{sides}: \\
-4 y^2+10 y-3 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(-3 x^2+10 x+\underline{\text{ }}\right)+\left(-4 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}-2 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-3 x^2+10 x+\underline{\text{ }}\right)=-3 \left(x^2-\frac{10 x}{3}+\underline{\text{ }}\right): \\
\fbox{$-3 \left(x^2-\frac{10 x}{3}+\underline{\text{ }}\right)$}+\left(-4 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}-2 \\
\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): \\
-3 \left(x^2-\frac{10 x}{3}+\underline{\text{ }}\right)+\fbox{$-4 \left(y^2-\frac{5 y}{2}+\underline{\text{ }}\right)$}=\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{-10}{3}}{2}\right)^2=\frac{25}{9} \text{on }\text{the }\text{left }\text{and }-3\times \frac{25}{9}=-\frac{25}{3} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-2-\frac{25}{3}=-\frac{31}{3}: \\
-3 \left(x^2-\frac{10 x}{3}+\frac{25}{9}\right)-4 \left(y^2-\frac{5 y}{2}+\underline{\text{ }}\right)=\fbox{$-\frac{31}{3}$} \\
\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{\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{31}{3}-\frac{25}{4}=-\frac{199}{12}: \\
-3 \left(x^2-\frac{10 x}{3}+\frac{25}{9}\right)-4 \left(y^2-\frac{5 y}{2}+\frac{25}{16}\right)=\fbox{$-\frac{199}{12}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{10 x}{3}+\frac{25}{9}=\left(x-\frac{5}{3}\right)^2: \\
-3 \fbox{$\left(x-\frac{5}{3}\right)^2$}-4 \left(y^2-\frac{5 y}{2}+\frac{25}{16}\right)=-\frac{199}{12} \\
\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{} & -3 \left(x-\frac{5}{3}\right)^2-4 \fbox{$\left(y-\frac{5}{4}\right)^2$}=-\frac{199}{12} \\
\end{array}
| khanacademy | amps |
Given the equation $7 x^2-3 x+5 y^2-6 y-1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
5 y^2-6 y+7 x^2-3 x-1=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }1 \text{to }\text{both }\text{sides}: \\
5 y^2-6 y+7 x^2-3 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(7 x^2-3 x+\underline{\text{ }}\right)+\left(5 y^2-6 y+\underline{\text{ }}\right)=\underline{\text{ }}+1 \\
\end{array}
Step 4:
\begin{array}{l}
\left(7 x^2-3 x+\underline{\text{ }}\right)=7 \left(x^2-\frac{3 x}{7}+\underline{\text{ }}\right): \\
\fbox{$7 \left(x^2-\frac{3 x}{7}+\underline{\text{ }}\right)$}+\left(5 y^2-6 y+\underline{\text{ }}\right)=\underline{\text{ }}+1 \\
\end{array}
Step 5:
\begin{array}{l}
\left(5 y^2-6 y+\underline{\text{ }}\right)=5 \left(y^2-\frac{6 y}{5}+\underline{\text{ }}\right): \\
7 \left(x^2-\frac{3 x}{7}+\underline{\text{ }}\right)+\fbox{$5 \left(y^2-\frac{6 y}{5}+\underline{\text{ }}\right)$}=\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 }7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-3}{7}}{2}\right)^2=\frac{9}{196} \text{on }\text{the }\text{left }\text{and }7\times \frac{9}{196}=\frac{9}{28} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
1+\frac{9}{28}=\frac{37}{28}: \\
7 \left(x^2-\frac{3 x}{7}+\frac{9}{196}\right)+5 \left(y^2-\frac{6 y}{5}+\underline{\text{ }}\right)=\fbox{$\frac{37}{28}$} \\
\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 }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 9:
\begin{array}{l}
\frac{37}{28}+\frac{9}{5}=\frac{437}{140}: \\
7 \left(x^2-\frac{3 x}{7}+\frac{9}{196}\right)+5 \left(y^2-\frac{6 y}{5}+\frac{9}{25}\right)=\fbox{$\frac{437}{140}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{3 x}{7}+\frac{9}{196}=\left(x-\frac{3}{14}\right)^2: \\
7 \fbox{$\left(x-\frac{3}{14}\right)^2$}+5 \left(y^2-\frac{6 y}{5}+\frac{9}{25}\right)=\frac{437}{140} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{6 y}{5}+\frac{9}{25}=\left(y-\frac{3}{5}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 7 \left(x-\frac{3}{14}\right)^2+5 \fbox{$\left(y-\frac{3}{5}\right)^2$}=\frac{437}{140} \\
\end{array}
| khanacademy | amps |
Given the equation $-10 x^2+5 x+4 y^2+6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-10 x^2+5 x+\left(4 y^2+6\right)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }4 y^2-10 x^2+5 x+6 \text{from }\text{both }\text{sides}: \\
10 x^2-5 x+\left(-4 y^2-6\right)=0 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Add }6 \text{to }\text{both }\text{sides}: \\
-4 y^2+10 x^2-5 x=6 \\
\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(10 x^2-5 x+\underline{\text{ }}\right)-4 y^2=\underline{\text{ }}+6 \\
\end{array}
Step 5:
\begin{array}{l}
\left(10 x^2-5 x+\underline{\text{ }}\right)=10 \left(x^2-\frac{x}{2}+\underline{\text{ }}\right): \\
\fbox{$10 \left(x^2-\frac{x}{2}+\underline{\text{ }}\right)$}-4 y^2=\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 }\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 7:
\begin{array}{l}
6+\frac{5}{8}=\frac{53}{8}: \\
10 \left(x^2-\frac{x}{2}+\frac{1}{16}\right)-4 y^2=\fbox{$\frac{53}{8}$} \\
\end{array}
Step 8:
\begin{array}{l}
x^2-\frac{x}{2}+\frac{1}{16}=\left(x-\frac{1}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & \text{10 }\fbox{$\left(x-\frac{1}{4}\right)^2$}-4 y^2=\frac{53}{8} \\
\end{array}
| khanacademy | amps |
Given the equation $8 x^2-4 x+y^2+7 y+5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
y^2+7 y+8 x^2-4 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^2-4 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(8 x^2-4 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^2-4 x+\underline{\text{ }}\right)=8 \left(x^2-\frac{x}{2}+\underline{\text{ }}\right): \\
\fbox{$8 \left(x^2-\frac{x}{2}+\underline{\text{ }}\right)$}+\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{-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 6:
\begin{array}{l}
\frac{1}{2}-5=-\frac{9}{2}: \\
8 \left(x^2-\frac{x}{2}+\frac{1}{16}\right)+\left(y^2+7 y+\underline{\text{ }}\right)=\fbox{$-\frac{9}{2}$} \\
\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{9}{2}=\frac{31}{4}: \\
8 \left(x^2-\frac{x}{2}+\frac{1}{16}\right)+\left(y^2+7 y+\frac{49}{4}\right)=\fbox{$\frac{31}{4}$} \\
\end{array}
Step 9:
\begin{array}{l}
x^2-\frac{x}{2}+\frac{1}{16}=\left(x-\frac{1}{4}\right)^2: \\
8 \fbox{$\left(x-\frac{1}{4}\right)^2$}+\left(y^2+7 y+\frac{49}{4}\right)=\frac{31}{4} \\
\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{1}{4}\right)^2+\fbox{$\left(y+\frac{7}{2}\right)^2$}=\frac{31}{4} \\
\end{array}
| khanacademy | amps |
Given the equation $3 x^2-8 y^2+y+8=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-8 y^2+y+\left(3 x^2+8\right)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }-8 y^2+y+3 x^2+8 \text{from }\text{both }\text{sides}: \\
8 y^2-y+\left(-3 x^2-8\right)=0 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Add }8 \text{to }\text{both }\text{sides}: \\
8 y^2-y-3 x^2=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(8 y^2-y+\underline{\text{ }}\right)-3 x^2=\underline{\text{ }}+8 \\
\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{$8 \left(y^2-\frac{y}{8}+\underline{\text{ }}\right)$}-3 x^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 }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}
8+\frac{1}{32}=\frac{257}{32}: \\
8 \left(y^2-\frac{y}{8}+\frac{1}{256}\right)-3 x^2=\fbox{$\frac{257}{32}$} \\
\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{$\left(y-\frac{1}{16}\right)^2$}-3 x^2=\frac{257}{32} \\
\end{array}
| khanacademy | amps |
Given the equation $8 x^2-9 x-10 y^2-4 y-7=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-10 y^2-4 y+8 x^2-9 x-7=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }7 \text{to }\text{both }\text{sides}: \\
-10 y^2-4 y+8 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(8 x^2-9 x+\underline{\text{ }}\right)+\left(-10 y^2-4 y+\underline{\text{ }}\right)=\underline{\text{ }}+7 \\
\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(-10 y^2-4 y+\underline{\text{ }}\right)=\underline{\text{ }}+7 \\
\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): \\
8 \left(x^2-\frac{9 x}{8}+\underline{\text{ }}\right)+\fbox{$-10 \left(y^2+\frac{2 y}{5}+\underline{\text{ }}\right)$}=\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{-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}
7+\frac{81}{32}=\frac{305}{32}: \\
8 \left(x^2-\frac{9 x}{8}+\frac{81}{256}\right)-10 \left(y^2+\frac{2 y}{5}+\underline{\text{ }}\right)=\fbox{$\frac{305}{32}$} \\
\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{\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{305}{32}-\frac{2}{5}=\frac{1461}{160}: \\
8 \left(x^2-\frac{9 x}{8}+\frac{81}{256}\right)-10 \left(y^2+\frac{2 y}{5}+\frac{1}{25}\right)=\fbox{$\frac{1461}{160}$} \\
\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{$\left(x-\frac{9}{16}\right)^2$}-10 \left(y^2+\frac{2 y}{5}+\frac{1}{25}\right)=\frac{1461}{160} \\
\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{} & 8 \left(x-\frac{9}{16}\right)^2-\text{10 }\fbox{$\left(y+\frac{1}{5}\right)^2$}=\frac{1461}{160} \\
\end{array}
| khanacademy | amps |
Given the equation $9 x^2+8 x+6 y^2-5 y+7=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
6 y^2-5 y+9 x^2+8 x+7=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }7 \text{from }\text{both }\text{sides}: \\
6 y^2-5 y+9 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(9 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(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(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): \\
9 \left(x^2+\frac{8 x}{9}+\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 }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}
\frac{16}{9}-7=-\frac{47}{9}: \\
9 \left(x^2+\frac{8 x}{9}+\frac{16}{81}\right)+6 \left(y^2-\frac{5 y}{6}+\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 }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{25}{24}-\frac{47}{9}=-\frac{301}{72}: \\
9 \left(x^2+\frac{8 x}{9}+\frac{16}{81}\right)+6 \left(y^2-\frac{5 y}{6}+\frac{25}{144}\right)=\fbox{$-\frac{301}{72}$} \\
\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{$\left(x+\frac{4}{9}\right)^2$}+6 \left(y^2-\frac{5 y}{6}+\frac{25}{144}\right)=-\frac{301}{72} \\
\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{} & 9 \left(x+\frac{4}{9}\right)^2+6 \fbox{$\left(y-\frac{5}{12}\right)^2$}=-\frac{301}{72} \\
\end{array}
| khanacademy | amps |
Given the equation $-5 x^2-4 x-2 y^2+10 y+5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-2 y^2+10 y-5 x^2-4 x+5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }5 \text{from }\text{both }\text{sides}: \\
-2 y^2+10 y-5 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(-5 x^2-4 x+\underline{\text{ }}\right)+\left(-2 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-5 x^2-4 x+\underline{\text{ }}\right)=-5 \left(x^2+\frac{4 x}{5}+\underline{\text{ }}\right): \\
\fbox{$-5 \left(x^2+\frac{4 x}{5}+\underline{\text{ }}\right)$}+\left(-2 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-2 y^2+10 y+\underline{\text{ }}\right)=-2 \left(y^2-5 y+\underline{\text{ }}\right): \\
-5 \left(x^2+\frac{4 x}{5}+\underline{\text{ }}\right)+\fbox{$-2 \left(y^2-5 y+\underline{\text{ }}\right)$}=\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{\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}
-5-\frac{4}{5}=-\frac{29}{5}: \\
-5 \left(x^2+\frac{4 x}{5}+\frac{4}{25}\right)-2 \left(y^2-5 y+\underline{\text{ }}\right)=\fbox{$-\frac{29}{5}$} \\
\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 }-2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-5}{2}\right)^2=\frac{25}{4} \text{on }\text{the }\text{left }\text{and }-2\times \frac{25}{4}=-\frac{25}{2} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{29}{5}-\frac{25}{2}=-\frac{183}{10}: \\
-5 \left(x^2+\frac{4 x}{5}+\frac{4}{25}\right)-2 \left(y^2-5 y+\frac{25}{4}\right)=\fbox{$-\frac{183}{10}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{4 x}{5}+\frac{4}{25}=\left(x+\frac{2}{5}\right)^2: \\
-5 \fbox{$\left(x+\frac{2}{5}\right)^2$}-2 \left(y^2-5 y+\frac{25}{4}\right)=-\frac{183}{10} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-5 y+\frac{25}{4}=\left(y-\frac{5}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -5 \left(x+\frac{2}{5}\right)^2-2 \fbox{$\left(y-\frac{5}{2}\right)^2$}=-\frac{183}{10} \\
\end{array}
| khanacademy | amps |
Given the equation $-7 x^2+2 x+5 y^2+6 y+10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
5 y^2+6 y-7 x^2+2 x+10=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }\text{10 }\text{from }\text{both }\text{sides}: \\
5 y^2+6 y-7 x^2+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(-7 x^2+2 x+\underline{\text{ }}\right)+\left(5 y^2+6 y+\underline{\text{ }}\right)=\underline{\text{ }}-10 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-7 x^2+2 x+\underline{\text{ }}\right)=-7 \left(x^2-\frac{2 x}{7}+\underline{\text{ }}\right): \\
\fbox{$-7 \left(x^2-\frac{2 x}{7}+\underline{\text{ }}\right)$}+\left(5 y^2+6 y+\underline{\text{ }}\right)=\underline{\text{ }}-10 \\
\end{array}
Step 5:
\begin{array}{l}
\left(5 y^2+6 y+\underline{\text{ }}\right)=5 \left(y^2+\frac{6 y}{5}+\underline{\text{ }}\right): \\
-7 \left(x^2-\frac{2 x}{7}+\underline{\text{ }}\right)+\fbox{$5 \left(y^2+\frac{6 y}{5}+\underline{\text{ }}\right)$}=\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 }-7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-2}{7}}{2}\right)^2=\frac{1}{49} \text{on }\text{the }\text{left }\text{and }\frac{-7}{49}=-\frac{1}{7} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-10-\frac{1}{7}=-\frac{71}{7}: \\
-7 \left(x^2-\frac{2 x}{7}+\frac{1}{49}\right)+5 \left(y^2+\frac{6 y}{5}+\underline{\text{ }}\right)=\fbox{$-\frac{71}{7}$} \\
\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{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 9:
\begin{array}{l}
\frac{9}{5}-\frac{71}{7}=-\frac{292}{35}: \\
-7 \left(x^2-\frac{2 x}{7}+\frac{1}{49}\right)+5 \left(y^2+\frac{6 y}{5}+\frac{9}{25}\right)=\fbox{$-\frac{292}{35}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{2 x}{7}+\frac{1}{49}=\left(x-\frac{1}{7}\right)^2: \\
-7 \fbox{$\left(x-\frac{1}{7}\right)^2$}+5 \left(y^2+\frac{6 y}{5}+\frac{9}{25}\right)=-\frac{292}{35} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{6 y}{5}+\frac{9}{25}=\left(y+\frac{3}{5}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -7 \left(x-\frac{1}{7}\right)^2+5 \fbox{$\left(y+\frac{3}{5}\right)^2$}=-\frac{292}{35} \\
\end{array}
| khanacademy | amps |
Given the equation $10 x^2+9 x-6 y^2-3 y+1=0$, 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+1=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }1 \text{from }\text{both }\text{sides}: \\
-6 y^2-3 y+10 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(10 x^2+9 x+\underline{\text{ }}\right)+\left(-6 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\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{ }}-1 \\
\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{ }}-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 }\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}-1=\frac{41}{40}: \\
10 \left(x^2+\frac{9 x}{10}+\frac{81}{400}\right)-6 \left(y^2+\frac{y}{2}+\underline{\text{ }}\right)=\fbox{$\frac{41}{40}$} \\
\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{41}{40}-\frac{3}{8}=\frac{13}{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{$\frac{13}{20}$} \\
\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{13}{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{13}{20} \\
\end{array}
| khanacademy | amps |
Given the equation $5 x^2-8 x+2 y^2-8 y-9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
2 y^2-8 y+5 x^2-8 x-9=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }9 \text{to }\text{both }\text{sides}: \\
2 y^2-8 y+5 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(5 x^2-8 x+\underline{\text{ }}\right)+\left(2 y^2-8 y+\underline{\text{ }}\right)=\underline{\text{ }}+9 \\
\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(2 y^2-8 y+\underline{\text{ }}\right)=\underline{\text{ }}+9 \\
\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{8 x}{5}+\underline{\text{ }}\right)+\fbox{$2 \left(y^2-4 y+\underline{\text{ }}\right)$}=\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 }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}
9+\frac{16}{5}=\frac{61}{5}: \\
5 \left(x^2-\frac{8 x}{5}+\frac{16}{25}\right)+2 \left(y^2-4 y+\underline{\text{ }}\right)=\fbox{$\frac{61}{5}$} \\
\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 }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{61}{5}+8=\frac{101}{5}: \\
5 \left(x^2-\frac{8 x}{5}+\frac{16}{25}\right)+2 \left(y^2-4 y+4\right)=\fbox{$\frac{101}{5}$} \\
\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{$\left(x-\frac{4}{5}\right)^2$}+2 \left(y^2-4 y+4\right)=\frac{101}{5} \\
\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{4}{5}\right)^2+2 \fbox{$(y-2)^2$}=\frac{101}{5} \\
\end{array}
| khanacademy | amps |
Given the equation $-10 x^2-5 y^2-8 y+9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-5 y^2-8 y+\left(9-10 x^2\right)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }-5 y^2-8 y-10 x^2+9 \text{from }\text{both }\text{sides}: \\
5 y^2+8 y+\left(10 x^2-9\right)=0 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Add }9 \text{to }\text{both }\text{sides}: \\
5 y^2+8 y+10 x^2=9 \\
\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 y^2+8 y+\underline{\text{ }}\right)+10 x^2=\underline{\text{ }}+9 \\
\end{array}
Step 5:
\begin{array}{l}
\left(5 y^2+8 y+\underline{\text{ }}\right)=5 \left(y^2+\frac{8 y}{5}+\underline{\text{ }}\right): \\
\fbox{$5 \left(y^2+\frac{8 y}{5}+\underline{\text{ }}\right)$}+10 x^2=\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 }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{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}
9+\frac{16}{5}=\frac{61}{5}: \\
5 \left(y^2+\frac{8 y}{5}+\frac{16}{25}\right)+10 x^2=\fbox{$\frac{61}{5}$} \\
\end{array}
Step 8:
\begin{array}{l}
y^2+\frac{8 y}{5}+\frac{16}{25}=\left(y+\frac{4}{5}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 5 \fbox{$\left(y+\frac{4}{5}\right)^2$}+10 x^2=\frac{61}{5} \\
\end{array}
| khanacademy | amps |
Given the equation $x^2-3 x+4 y^2-10 y+9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
4 y^2-10 y+x^2-3 x+9=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }9 \text{from }\text{both }\text{sides}: \\
4 y^2-10 y+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(x^2-3 x+\underline{\text{ }}\right)+\left(4 y^2-10 y+\underline{\text{ }}\right)=\underline{\text{ }}-9 \\
\end{array}
Step 4:
\begin{array}{l}
\left(4 y^2-10 y+\underline{\text{ }}\right)=4 \left(y^2-\frac{5 y}{2}+\underline{\text{ }}\right): \\
\left(x^2-3 x+\underline{\text{ }}\right)+\fbox{$4 \left(y^2-\frac{5 y}{2}+\underline{\text{ }}\right)$}=\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 }x \text{and }\text{square }\text{it, }\text{then }\text{add }\text{it }\text{to }\text{both }\text{sides.} \\
\text{Add }\left(\frac{-3}{2}\right)^2=\frac{9}{4} \text{to }\text{both }\text{sides}: \\
\end{array}
Step 6:
\begin{array}{l}
\frac{9}{4}-9=-\frac{27}{4}: \\
\left(x^2-3 x+\frac{9}{4}\right)+4 \left(y^2-\frac{5 y}{2}+\underline{\text{ }}\right)=\fbox{$-\frac{27}{4}$} \\
\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{-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 8:
\begin{array}{l}
\frac{25}{4}-\frac{27}{4}=-\frac{1}{2}: \\
\left(x^2-3 x+\frac{9}{4}\right)+4 \left(y^2-\frac{5 y}{2}+\frac{25}{16}\right)=\fbox{$-\frac{1}{2}$} \\
\end{array}
Step 9:
\begin{array}{l}
x^2-3 x+\frac{9}{4}=\left(x-\frac{3}{2}\right)^2: \\
\fbox{$\left(x-\frac{3}{2}\right)^2$}+4 \left(y^2-\frac{5 y}{2}+\frac{25}{16}\right)=-\frac{1}{2} \\
\end{array}
Step 10:
\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{} & \left(x-\frac{3}{2}\right)^2+4 \fbox{$\left(y-\frac{5}{4}\right)^2$}=-\frac{1}{2} \\
\end{array}
| khanacademy | amps |
Given the equation $-10 x^2-3 x+9 y^2-6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-10 x^2-3 x+\left(9 y^2-6\right)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }9 y^2-10 x^2-3 x-6 \text{from }\text{both }\text{sides}: \\
10 x^2+3 x+\left(6-9 y^2\right)=0 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Subtract }6 \text{from }\text{both }\text{sides}: \\
-9 y^2+10 x^2+3 x=-6 \\
\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(10 x^2+3 x+\underline{\text{ }}\right)-9 y^2=\underline{\text{ }}-6 \\
\end{array}
Step 5:
\begin{array}{l}
\left(10 x^2+3 x+\underline{\text{ }}\right)=10 \left(x^2+\frac{3 x}{10}+\underline{\text{ }}\right): \\
\fbox{$10 \left(x^2+\frac{3 x}{10}+\underline{\text{ }}\right)$}-9 y^2=\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 }\text{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}
\frac{9}{40}-6=-\frac{231}{40}: \\
10 \left(x^2+\frac{3 x}{10}+\frac{9}{400}\right)-9 y^2=\fbox{$-\frac{231}{40}$} \\
\end{array}
Step 8:
\begin{array}{l}
x^2+\frac{3 x}{10}+\frac{9}{400}=\left(x+\frac{3}{20}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & \text{10 }\fbox{$\left(x+\frac{3}{20}\right)^2$}-9 y^2=-\frac{231}{40} \\
\end{array}
| khanacademy | amps |
Given the equation $-8 x^2-6 x-6 y^2-6 y+5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-6 y^2-6 y-8 x^2-6 x+5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }5 \text{from }\text{both }\text{sides}: \\
-6 y^2-6 y-8 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(-8 x^2-6 x+\underline{\text{ }}\right)+\left(-6 y^2-6 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\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(-6 y^2-6 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\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{3 x}{4}+\underline{\text{ }}\right)+\fbox{$-6 \left(y^2+y+\underline{\text{ }}\right)$}=\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 }-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}
-5-\frac{9}{8}=-\frac{49}{8}: \\
-8 \left(x^2+\frac{3 x}{4}+\frac{9}{64}\right)-6 \left(y^2+y+\underline{\text{ }}\right)=\fbox{$-\frac{49}{8}$} \\
\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{49}{8}-\frac{3}{2}=-\frac{61}{8}: \\
-8 \left(x^2+\frac{3 x}{4}+\frac{9}{64}\right)-6 \left(y^2+y+\frac{1}{4}\right)=\fbox{$-\frac{61}{8}$} \\
\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{$\left(x+\frac{3}{8}\right)^2$}-6 \left(y^2+y+\frac{1}{4}\right)=-\frac{61}{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{3}{8}\right)^2-6 \fbox{$\left(y+\frac{1}{2}\right)^2$}=-\frac{61}{8} \\
\end{array}
| khanacademy | amps |
Given the equation $5 x^2-3 x+2 y+9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
5 x^2-3 x+(2 y+9)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }2 y+9 \text{from }\text{both }\text{sides}: \\
5 x^2-3 x=-2 y-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(5 x^2-3 x+\underline{\text{ }}\right)=(-2 y-9)+\underline{\text{ }} \\
\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{$5 \left(x^2-\frac{3 x}{5}+\underline{\text{ }}\right)$}=(-2 y-9)+\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 }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 6:
\begin{array}{l}
(-2 y-9)+\frac{9}{20}=-2 y-\frac{171}{20}: \\
5 \left(x^2-\frac{3 x}{5}+\frac{9}{100}\right)=\fbox{$-2 y-\frac{171}{20}$} \\
\end{array}
Step 7:
\begin{array}{l}
x^2-\frac{3 x}{5}+\frac{9}{100}=\left(x-\frac{3}{10}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 5 \fbox{$\left(x-\frac{3}{10}\right)^2$}=-2 y-\frac{171}{20} \\
\end{array}
| khanacademy | amps |
Given the equation $-9 x^2+4 x+3 y^2-9 y+10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
3 y^2-9 y-9 x^2+4 x+10=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }\text{10 }\text{from }\text{both }\text{sides}: \\
3 y^2-9 y-9 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(-9 x^2+4 x+\underline{\text{ }}\right)+\left(3 y^2-9 y+\underline{\text{ }}\right)=\underline{\text{ }}-10 \\
\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(3 y^2-9 y+\underline{\text{ }}\right)=\underline{\text{ }}-10 \\
\end{array}
Step 5:
\begin{array}{l}
\left(3 y^2-9 y+\underline{\text{ }}\right)=3 \left(y^2-3 y+\underline{\text{ }}\right): \\
-9 \left(x^2-\frac{4 x}{9}+\underline{\text{ }}\right)+\fbox{$3 \left(y^2-3 y+\underline{\text{ }}\right)$}=\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 }-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}
-10-\frac{4}{9}=-\frac{94}{9}: \\
-9 \left(x^2-\frac{4 x}{9}+\frac{4}{81}\right)+3 \left(y^2-3 y+\underline{\text{ }}\right)=\fbox{$-\frac{94}{9}$} \\
\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 }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 9:
\begin{array}{l}
\frac{27}{4}-\frac{94}{9}=-\frac{133}{36}: \\
-9 \left(x^2-\frac{4 x}{9}+\frac{4}{81}\right)+3 \left(y^2-3 y+\frac{9}{4}\right)=\fbox{$-\frac{133}{36}$} \\
\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{$\left(x-\frac{2}{9}\right)^2$}+3 \left(y^2-3 y+\frac{9}{4}\right)=-\frac{133}{36} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-3 y+\frac{9}{4}=\left(y-\frac{3}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -9 \left(x-\frac{2}{9}\right)^2+3 \fbox{$\left(y-\frac{3}{2}\right)^2$}=-\frac{133}{36} \\
\end{array}
| khanacademy | amps |
Given the equation $-9 x^2+8 x-y^2-5 y-1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-y^2-5 y-9 x^2+8 x-1=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }1 \text{to }\text{both }\text{sides}: \\
-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(-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(-y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}+1 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-y^2-5 y+\underline{\text{ }}\right)=-\left(y^2+5 y+\underline{\text{ }}\right): \\
-9 \left(x^2-\frac{8 x}{9}+\underline{\text{ }}\right)+\fbox{$-\left(y^2+5 y+\underline{\text{ }}\right)$}=\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 }-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{7}{9}: \\
-9 \left(x^2-\frac{8 x}{9}+\frac{16}{81}\right)-\left(y^2+5 y+\underline{\text{ }}\right)=\fbox{$-\frac{7}{9}$} \\
\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 }-1 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{5}{2}\right)^2=\frac{25}{4} \text{on }\text{the }\text{left }\text{and }-\frac{25}{4}=-\frac{25}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{7}{9}-\frac{25}{4}=-\frac{253}{36}: \\
-9 \left(x^2-\frac{8 x}{9}+\frac{16}{81}\right)-\left(y^2+5 y+\frac{25}{4}\right)=\fbox{$-\frac{253}{36}$} \\
\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{$\left(x-\frac{4}{9}\right)^2$}-\left(y^2+5 y+\frac{25}{4}\right)=-\frac{253}{36} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+5 y+\frac{25}{4}=\left(y+\frac{5}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -9 \left(x-\frac{4}{9}\right)^2-\fbox{$\left(y+\frac{5}{2}\right)^2$}=-\frac{253}{36} \\
\end{array}
| khanacademy | amps |
Given the equation $7 x^2-8 x+9 y^2-3 y+2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
9 y^2-3 y+7 x^2-8 x+2=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }2 \text{from }\text{both }\text{sides}: \\
9 y^2-3 y+7 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(7 x^2-8 x+\underline{\text{ }}\right)+\left(9 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}-2 \\
\end{array}
Step 4:
\begin{array}{l}
\left(7 x^2-8 x+\underline{\text{ }}\right)=7 \left(x^2-\frac{8 x}{7}+\underline{\text{ }}\right): \\
\fbox{$7 \left(x^2-\frac{8 x}{7}+\underline{\text{ }}\right)$}+\left(9 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}-2 \\
\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): \\
7 \left(x^2-\frac{8 x}{7}+\underline{\text{ }}\right)+\fbox{$9 \left(y^2-\frac{y}{3}+\underline{\text{ }}\right)$}=\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 }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 7:
\begin{array}{l}
\frac{16}{7}-2=\frac{2}{7}: \\
7 \left(x^2-\frac{8 x}{7}+\frac{16}{49}\right)+9 \left(y^2-\frac{y}{3}+\underline{\text{ }}\right)=\fbox{$\frac{2}{7}$} \\
\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 }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{2}{7}+\frac{1}{4}=\frac{15}{28}: \\
7 \left(x^2-\frac{8 x}{7}+\frac{16}{49}\right)+9 \left(y^2-\frac{y}{3}+\frac{1}{36}\right)=\fbox{$\frac{15}{28}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{8 x}{7}+\frac{16}{49}=\left(x-\frac{4}{7}\right)^2: \\
7 \fbox{$\left(x-\frac{4}{7}\right)^2$}+9 \left(y^2-\frac{y}{3}+\frac{1}{36}\right)=\frac{15}{28} \\
\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{} & 7 \left(x-\frac{4}{7}\right)^2+9 \fbox{$\left(y-\frac{1}{6}\right)^2$}=\frac{15}{28} \\
\end{array}
| khanacademy | amps |
Given the equation $6 x^2+9 x+y^2+9 y-6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
y^2+9 y+6 x^2+9 x-6=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }6 \text{to }\text{both }\text{sides}: \\
y^2+9 y+6 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(6 x^2+9 x+\underline{\text{ }}\right)+\left(y^2+9 y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\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(y^2+9 y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\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 }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 6:
\begin{array}{l}
6+\frac{27}{8}=\frac{75}{8}: \\
6 \left(x^2+\frac{3 x}{2}+\frac{9}{16}\right)+\left(y^2+9 y+\underline{\text{ }}\right)=\fbox{$\frac{75}{8}$} \\
\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{9}{2}\right)^2=\frac{81}{4} \text{to }\text{both }\text{sides}: \\
\end{array}
Step 8:
\begin{array}{l}
\frac{75}{8}+\frac{81}{4}=\frac{237}{8}: \\
6 \left(x^2+\frac{3 x}{2}+\frac{9}{16}\right)+\left(y^2+9 y+\frac{81}{4}\right)=\fbox{$\frac{237}{8}$} \\
\end{array}
Step 9:
\begin{array}{l}
x^2+\frac{3 x}{2}+\frac{9}{16}=\left(x+\frac{3}{4}\right)^2: \\
6 \fbox{$\left(x+\frac{3}{4}\right)^2$}+\left(y^2+9 y+\frac{81}{4}\right)=\frac{237}{8} \\
\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{} & 6 \left(x+\frac{3}{4}\right)^2+\fbox{$\left(y+\frac{9}{2}\right)^2$}=\frac{237}{8} \\
\end{array}
| khanacademy | amps |
Given the equation $-9 x^2-6 x+6 y^2-7 y+6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
6 y^2-7 y-9 x^2-6 x+6=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }6 \text{from }\text{both }\text{sides}: \\
6 y^2-7 y-9 x^2-6 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^2-6 x+\underline{\text{ }}\right)+\left(6 y^2-7 y+\underline{\text{ }}\right)=\underline{\text{ }}-6 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-9 x^2-6 x+\underline{\text{ }}\right)=-9 \left(x^2+\frac{2 x}{3}+\underline{\text{ }}\right): \\
\fbox{$-9 \left(x^2+\frac{2 x}{3}+\underline{\text{ }}\right)$}+\left(6 y^2-7 y+\underline{\text{ }}\right)=\underline{\text{ }}-6 \\
\end{array}
Step 5:
\begin{array}{l}
\left(6 y^2-7 y+\underline{\text{ }}\right)=6 \left(y^2-\frac{7 y}{6}+\underline{\text{ }}\right): \\
-9 \left(x^2+\frac{2 x}{3}+\underline{\text{ }}\right)+\fbox{$6 \left(y^2-\frac{7 y}{6}+\underline{\text{ }}\right)$}=\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{2}{3}}{2}\right)^2=\frac{1}{9} \text{on }\text{the }\text{left }\text{and }\frac{-9}{9}=-1 \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-6-1=-7: \\
-9 \left(x^2+\frac{2 x}{3}+\frac{1}{9}\right)+6 \left(y^2-\frac{7 y}{6}+\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 }6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-7}{6}}{2}\right)^2=\frac{49}{144} \text{on }\text{the }\text{left }\text{and }6\times \frac{49}{144}=\frac{49}{24} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{49}{24}-7=-\frac{119}{24}: \\
-9 \left(x^2+\frac{2 x}{3}+\frac{1}{9}\right)+6 \left(y^2-\frac{7 y}{6}+\frac{49}{144}\right)=\fbox{$-\frac{119}{24}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{2 x}{3}+\frac{1}{9}=\left(x+\frac{1}{3}\right)^2: \\
-9 \fbox{$\left(x+\frac{1}{3}\right)^2$}+6 \left(y^2-\frac{7 y}{6}+\frac{49}{144}\right)=-\frac{119}{24} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{7 y}{6}+\frac{49}{144}=\left(y-\frac{7}{12}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -9 \left(x+\frac{1}{3}\right)^2+6 \fbox{$\left(y-\frac{7}{12}\right)^2$}=-\frac{119}{24} \\
\end{array}
| khanacademy | amps |
Given the equation $7 x^2+7 x-5 y^2-6 y-5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-5 y^2-6 y+7 x^2+7 x-5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }5 \text{to }\text{both }\text{sides}: \\
-5 y^2-6 y+7 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(7 x^2+7 x+\underline{\text{ }}\right)+\left(-5 y^2-6 y+\underline{\text{ }}\right)=\underline{\text{ }}+5 \\
\end{array}
Step 4:
\begin{array}{l}
\left(7 x^2+7 x+\underline{\text{ }}\right)=7 \left(x^2+x+\underline{\text{ }}\right): \\
\fbox{$7 \left(x^2+x+\underline{\text{ }}\right)$}+\left(-5 y^2-6 y+\underline{\text{ }}\right)=\underline{\text{ }}+5 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-5 y^2-6 y+\underline{\text{ }}\right)=-5 \left(y^2+\frac{6 y}{5}+\underline{\text{ }}\right): \\
7 \left(x^2+x+\underline{\text{ }}\right)+\fbox{$-5 \left(y^2+\frac{6 y}{5}+\underline{\text{ }}\right)$}=\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{1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{7}{4}=\frac{7}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
5+\frac{7}{4}=\frac{27}{4}: \\
7 \left(x^2+x+\frac{1}{4}\right)-5 \left(y^2+\frac{6 y}{5}+\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 }-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 9:
\begin{array}{l}
\frac{27}{4}-\frac{9}{5}=\frac{99}{20}: \\
7 \left(x^2+x+\frac{1}{4}\right)-5 \left(y^2+\frac{6 y}{5}+\frac{9}{25}\right)=\fbox{$\frac{99}{20}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+x+\frac{1}{4}=\left(x+\frac{1}{2}\right)^2: \\
7 \fbox{$\left(x+\frac{1}{2}\right)^2$}-5 \left(y^2+\frac{6 y}{5}+\frac{9}{25}\right)=\frac{99}{20} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{6 y}{5}+\frac{9}{25}=\left(y+\frac{3}{5}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 7 \left(x+\frac{1}{2}\right)^2-5 \fbox{$\left(y+\frac{3}{5}\right)^2$}=\frac{99}{20} \\
\end{array}
| khanacademy | amps |
Given the equation $x^2-9 x+2 y^2-5 y-5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
2 y^2-5 y+x^2-9 x-5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }5 \text{to }\text{both }\text{sides}: \\
2 y^2-5 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(2 y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}+5 \\
\end{array}
Step 4:
\begin{array}{l}
\left(2 y^2-5 y+\underline{\text{ }}\right)=2 \left(y^2-\frac{5 y}{2}+\underline{\text{ }}\right): \\
\left(x^2-9 x+\underline{\text{ }}\right)+\fbox{$2 \left(y^2-\frac{5 y}{2}+\underline{\text{ }}\right)$}=\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{-9}{2}\right)^2=\frac{81}{4} \text{to }\text{both }\text{sides}: \\
\end{array}
Step 6:
\begin{array}{l}
5+\frac{81}{4}=\frac{101}{4}: \\
\left(x^2-9 x+\frac{81}{4}\right)+2 \left(y^2-\frac{5 y}{2}+\underline{\text{ }}\right)=\fbox{$\frac{101}{4}$} \\
\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 }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 8:
\begin{array}{l}
\frac{101}{4}+\frac{25}{8}=\frac{227}{8}: \\
\left(x^2-9 x+\frac{81}{4}\right)+2 \left(y^2-\frac{5 y}{2}+\frac{25}{16}\right)=\fbox{$\frac{227}{8}$} \\
\end{array}
Step 9:
\begin{array}{l}
x^2-9 x+\frac{81}{4}=\left(x-\frac{9}{2}\right)^2: \\
\fbox{$\left(x-\frac{9}{2}\right)^2$}+2 \left(y^2-\frac{5 y}{2}+\frac{25}{16}\right)=\frac{227}{8} \\
\end{array}
Step 10:
\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{} & \left(x-\frac{9}{2}\right)^2+2 \fbox{$\left(y-\frac{5}{4}\right)^2$}=\frac{227}{8} \\
\end{array}
| khanacademy | amps |