Answer
The solution set of equation $\sqrt{8-2x}-x=0$ is $\left\{ \left. 2 \right\} \right.$.
Work Step by Step
Consider the provided equation, $\sqrt{8-2x}-x=0$
The above equation can be written as:
$\begin{align}
& \sqrt{8-2x}-x=0 \\
& \sqrt{8-2x}=x
\end{align}$
Square both sides
$\begin{align}
& 8-2x={{x}^{2}} \\
& {{x}^{2}}+2x-8=0
\end{align}$
Apply quadratic formula
$x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Here a is 1, b is $2$ and c is $-8$
$\begin{align}
& x=\frac{-2\pm \sqrt{{{\left( -2 \right)}^{2}}-4\left( 1 \right)\left( -8 \right)}}{2\left( 1 \right)} \\
& =\frac{-2\pm \sqrt{4+32}}{2} \\
& =\frac{-2\pm \sqrt{36}}{2} \\
& =\frac{-2\pm 6}{2}
\end{align}$
On further simplification
$\begin{align}
& x=\frac{-2+6}{2} \\
& =\frac{4}{2} \\
& =2
\end{align}$
Or
$\begin{align}
& x=\frac{-2-6}{2} \\
& =\frac{-8}{2} \\
& =-4
\end{align}$
Check:
For $x=2$
Put $x=2$ in $\sqrt{8-2x}-x=0$
$\begin{align}
& \sqrt{8-2\left( 2 \right)}-2\overset{?}{\mathop{=}}\,0 \\
& \sqrt{8-4}-2\overset{?}{\mathop{=}}\,0 \\
& \sqrt{4}-2\overset{?}{\mathop{=}}\,0 \\
& 2-2\overset{?}{\mathop{=}}\,0
\end{align}$
Which is true.
For $x=-4$
Put $x=-4$ in $\sqrt{8-2x}-x=0$
$\begin{align}
& \sqrt{8-2\left( -4 \right)}-\left( -4 \right)\overset{?}{\mathop{=}}\,4 \\
& \sqrt{8+8}+4\overset{?}{\mathop{=}}\,4 \\
& \sqrt{16}+4\overset{?}{\mathop{=}}\,4 \\
& 4+4\overset{?}{\mathop{=}}\,4
\end{align}$
Therefore
$8\overset{?}{\mathop{=}}\,4$
Which is not true.
Therefore, the solution set of the equation $\sqrt{8-2x}-x=0$ is $\left\{ \left. 2 \right\} \right.$.
