Answer
$\left\{3,11\right\}$
Work Step by Step
Squaring both sides, the given equation, $
\sqrt{2x+3}=2+\sqrt{x-2}
,$ is equivalent to
\begin{align*}
\left(\sqrt{2x+3}\right)^2&=\left(2+\sqrt{x-2}\right)^2
\\
2x+3&=\left(2+\sqrt{x-2}\right)^2
\\
2x+3&=(2)^2+2(2)\left(\sqrt{x-2}\right)+\left(\sqrt{x-2}\right)^2
&(\text{use }(a+b)^2=a^2+2ab+b^2)
\\
2x+3&=4+4(\sqrt{x-2})+x-2
\\
(2x-x)+(3-4+2)&=4(\sqrt{x-2})
\\
x+1&=4\sqrt{x-2}
.\end{align*}
Squaring both sides again, the equation above is equivalent to
\begin{align*}
(x+1)^2&=\left(4\sqrt{x-2}\right)^2
\\
(x)^2+2(x)(1)+(1)^2&=\left(4\sqrt{x-2}\right)^2
&(\text{use }(a+b)^2=a^2+2ab+b^2)
\\
x^2+2x+1&=16\left(x-2\right)
\\
x^2+2x+1&=16x-32
\\
x^2+(2x-16x)+(1+32)&=0
\\
x^2-14x+33&=0
.\end{align*}
Using factoring of trinomials, the equation above is equivalent to
\begin{align*}
(x-11)(x-3)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving for the variable, then
\begin{array}{l|r}
x-11=0 & x-3=0
\\
x=11 & x=3
\end{array}
Checking by substituting the solutions in the given equation results to
\begin{array}{l|r}
\text{If }x=11: & \text{If }x=3:
\\\\
\sqrt{2(11)+3}\overset{?}=2+\sqrt{11-2} &
\sqrt{2(3)+3}\overset{?}=2+\sqrt{3-2}
\\
\sqrt{22+3}\overset{?}=2+\sqrt{9} &
\sqrt{6+3}\overset{?}=2+\sqrt{1}
\\
\sqrt{25}\overset{?}=2+3 &
\sqrt{9}\overset{?}=2+1
\\
5\overset{\checkmark}=5 &
3\overset{\checkmark}=3
.\end{array}
Since all solutions satisfy the given equation, then the solution set of the equation $
\sqrt{2x+3}=2+\sqrt{x-2}
$ is $\left\{3,11\right\}$.
