Answer
The solution set is$\underline{\left\{ 22 \right\}}$.
Work Step by Step
$\sqrt{2x+5}-\sqrt{x+3}=2$
Add$\sqrt{x+3}$ to both sides:
$\sqrt{2x+5}-\sqrt{x+3}+\sqrt{x+3}=2+\sqrt{x+3}$
Simplify:
$\sqrt{2x+5}=2+\sqrt{x+3}$
Square both the sides:
${{\left( \sqrt{2x+5} \right)}^{2}}={{\left( 2+\sqrt{x+3} \right)}^{2}}$
Simplify, square eliminates the square root and using ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$.
$\begin{align}
& 2x+5={{2}^{2}}+2\times 2\times \sqrt{x+3}+{{\left( \sqrt{x+3} \right)}^{2}} \\
& 2x+5=4+4\sqrt{x+3}+x+3 \\
& 2x+5=7+4\sqrt{x+3}+x
\end{align}$
Subtract$7$ from both sides:
$2x+5-7=7-7+4\sqrt{x+3}+x$
Simplify:
$2x-2=4\sqrt{x+3}+x$
Subtract x from both sides:
$2x-x-2=4\sqrt{x+3}+x-x$
Simplify:
$x-2=4\sqrt{x+3}$
Again, square both the sides:
${{\left( x-2 \right)}^{2}}={{\left( 4\sqrt{x+3} \right)}^{2}}$
Simplify, square eliminates the square root and using${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$.
${{x}^{2}}-2\times x\times 2+4=16\left( x+3 \right)$
.
Use distributive,
${{x}^{2}}-4x+4=16x+48$
Subtract$16x+48$ from both sides:
\[\begin{align}
& {{x}^{2}}-4x+4-\left( 16x+48 \right)=16x+48-\left( 16x+48 \right) \\
& {{x}^{2}}-4x+4-16x-48=16x+48-16x-48 \\
& {{x}^{2}}-20x-44=0
\end{align}\]
Factorize,
$\begin{align}
& {{x}^{2}}-22x+2x-44=0 \\
& x\left( x-22 \right)+2\left( x-22 \right)=0 \\
& \left( x-22 \right)\left( x+2 \right)=0
\end{align}$
Compare each factor separately:
\[\begin{align}
& x-22=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x+2=0 \\
& x=22\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=-2 \\
\end{align}\]
Check: $x=22$
$\sqrt{2x+5}-\sqrt{x+3}=2$
Substitute and proceed from L.H.S:
$\begin{align}
& \sqrt{2\left( 22 \right)+5}-\sqrt{22+3} \\
& \sqrt{44+5}-\sqrt{25} \\
& \sqrt{49}-\sqrt{25} \\
& 7-5 \\
& 2 \\
\end{align}$
which is true.
Check: $x=-2$
$\sqrt{2x+5}-\sqrt{x+3}=2$
Substitute and proceed from L.H.S
$\begin{align}
& \sqrt{2\left( -2 \right)+5}-\sqrt{-2+3} \\
& \sqrt{-4+5}-\sqrt{1} \\
& \sqrt{1}-1 \\
& 1-1 \\
& 0 \\
\end{align}$
which is not true.
Hence, the solution is $x=22$only.
Hence, the solution set is$\left\{ 22 \right\}$.
