Answer
The solution of the quadratic equation $0=-2{{\left( x-3 \right)}^{2}}+8$ is $\left\{ 1,5 \right\}$.
Work Step by Step
Consider the quadratic equation,
$0=-2{{\left( x-3 \right)}^{2}}+8$
Add the term $2{{\left( x-3 \right)}^{2}}$ on both sides,
$\begin{align}
& 0+2{{\left( x-3 \right)}^{2}}=-2{{\left( x-3 \right)}^{2}}+8+2{{\left( x-3 \right)}^{2}} \\
& 2{{\left( x-3 \right)}^{2}}=8
\end{align}$
Divide both sides by 2,
$\begin{align}
& 0+2{{\left( x-3 \right)}^{2}}=-2{{\left( x-3 \right)}^{2}}+8+2{{\left( x-3 \right)}^{2}} \\
& \frac{2}{2}{{\left( x-3 \right)}^{2}}=\frac{8}{2} \\
& {{\left( x-3 \right)}^{2}}=4
\end{align}$
Apply the square root property,
$\begin{align}
& x-3=\pm \sqrt{4} \\
& x-3=\pm 2 \\
& x=3\pm 2
\end{align}$
First take the positive sign,
$\begin{align}
& x=3+2 \\
& =5
\end{align}$
Secondly, take the negative sign,
$\begin{align}
& x=3-2 \\
& =1
\end{align}$
Thus, the two values of x are 5 and 1.
Thus, the solutions of the quadratic equation $0=-2{{\left( x-3 \right)}^{2}}+8$ are 1 and 5. Check it by substituting the value of x in the equations.
Check:
For, $x=1$
\[\begin{align}
& 0\overset{?}{\mathop{=}}\,-2{{\left( 1-3 \right)}^{2}}+8 \\
& 0\overset{?}{\mathop{=}}\,-2{{\left( -2 \right)}^{2}}+8 \\
& 0\overset{?}{\mathop{=}}\,-8+8 \\
& 0\overset{?}{\mathop{=}}\,0
\end{align}\]
Which is true.
For, $x=5$
\[\begin{align}
& 0\overset{?}{\mathop{=}}\,-2{{\left( 5-3 \right)}^{2}}+8 \\
& 0\overset{?}{\mathop{=}}\,-2{{\left( 2 \right)}^{2}}+8 \\
& 0\overset{?}{\mathop{=}}\,-8+8 \\
& 0\overset{?}{\mathop{=}}\,0
\end{align}\]
Which is true.
Thus, the solution of the quadratic equation $0=-2{{\left( x-3 \right)}^{2}}+8$ is $\left\{ 1,5 \right\}$.
