Answer
The solution of the equation $\sqrt{2x-1}+2=x$ is $\left\{ 5 \right\}$ .
Work Step by Step
Consider the equation,
$\sqrt{2x-1}+2=x$
To isolate the radical to one side of the equation, subtract $2$ from both sides of the equation,
$\begin{align}
& \sqrt{2x-1}+2-2=x-2 \\
& \sqrt{2x-1}=x-2
\end{align}$
Square both sides of the equation to eliminate the radical. So,
${{\left( \sqrt{2x-1} \right)}^{2}}={{\left( x-2 \right)}^{2}}$
Use the formula
${{\left( A-B \right)}^{2}}={{A}^{2}}-2AB+{{B}^{2}}$
to expand the right side of the equation,
$2x-1={{x}^{2}}-4x+4$
Subtract $2x$ and add $1$ to form the equation in quadratic form. So,
$\begin{align}
& 2x-1-2x+1={{x}^{2}}-4x+4-2x+1 \\
& 0={{x}^{2}}-6x+5
\end{align}$
Factorize the quadratic equation,
$\begin{align}
& {{x}^{2}}-6x+5=0 \\
& \left( x-1 \right)\left( x-5 \right)=0 \\
& x=1\text{ or }x=5
\end{align}$
Now, verify the answer by substituting the obtained solutions into the original equation.
So, first substitute $x=1$ in the equation $\sqrt{2x-1}+2=x$. Therefore,
$\begin{align}
\sqrt{2\left( 1 \right)-1}+2\overset{?}{\mathop{=}}\,1 & \\
\sqrt{1}+2\overset{?}{\mathop{=}}\,1 & \\
1+2\overset{?}{\mathop{=}}\,1 & \\
3\overset{?}{\mathop{=}}\,1 & \\
\end{align}$
This is a false statement.
Hence, the solution of the equation cannot be $x=1$.
Now, substitute $x=5$ into the original equation $\sqrt{2x-1}+2=x$.
$\begin{align}
\sqrt{2\left( 5 \right)-1}+\overset{?}{\mathop{=}}\,5 & \\
\sqrt{9}+2\overset{?}{\mathop{=}}\,5 & \\
3+2\overset{?}{\mathop{=}}\,5 & \\
5=5 & \\
\end{align}$
This statement is true if the solution of the equation is $x=5$.
Hence, the solution set of the equation $\sqrt{2x-1}+2=x$ is $\left\{ 5 \right\}$.
