Answer
The solution set of equation $\sqrt{x-3}+5=x$ is $\left\{ \left. 7 \right\} \right.$.
Work Step by Step
Consider the provided equation, $\sqrt{x-3}+5=x$
The provided equation can be written as:
$\begin{align}
& \sqrt{x-3}+5=x \\
& \sqrt{x-3}=x-5
\end{align}$
Square both sides
$x-3={{\left( x-5 \right)}^{2}}$
Apply ${{\left( x-y \right)}^{2}}={{x}^{2}}+{{y}^{2}}-2xy$
$\begin{align}
& x-3={{x}^{2}}+25-10x \\
& {{x}^{2}}+25-10x-x+3=0 \\
& {{x}^{2}}-11x+28=0 \\
\end{align}$
Apply quadratic formula
$x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Here a is 1, b is $-11$ and c is $28$
$\begin{align}
& x=\frac{11\pm \sqrt{{{\left( -11 \right)}^{2}}-4\left( 1 \right)\left( 28 \right)}}{2\left( 1 \right)} \\
& =\frac{11\pm \sqrt{121-112}}{2} \\
& =\frac{11\pm 3}{2}
\end{align}$
On further simplification
$\begin{align}
& x=\frac{11+3}{2} \\
& =\frac{14}{2} \\
& =7
\end{align}$
Or
$\begin{align}
& x=\frac{11-3}{2} \\
& =\frac{8}{2} \\
& =4
\end{align}$
Check:
For $x=7$
Put $x=7$ in $\sqrt{x-3}+5=x$
$\begin{align}
& \sqrt{7-3}+5\overset{?}{\mathop{=}}\,7 \\
& \sqrt{4}+5\overset{?}{\mathop{=}}\,7 \\
& 2+5\overset{?}{\mathop{=}}\,7 \\
& 7\overset{?}{\mathop{=}}\,7
\end{align}$
Which is true.
For $x=4$
Put $x=4$ in $\sqrt{x-3}+5=x$
$\begin{align}
& \sqrt{4-3}+5\overset{?}{\mathop{=}}\,4 \\
& \sqrt{1}+5\overset{?}{\mathop{=}}\,4 \\
& 1+5\overset{?}{\mathop{=}}\,4 \\
& 6\overset{?}{\mathop{=}}\,4
\end{align}$
Which is not true.
Therefore, the solution set of the equation $\sqrt{x-3}+5=x$ is $\left\{ \left. 7 \right\} \right.$.
