Answer
$\{0,4\}$
Work Step by Step
Squaring both sides of the given equation, $
\sqrt{2x+1}-\sqrt{x}=1
$, results to
\begin{align*}
\left(\sqrt{2x+1}-\sqrt{x}\right)^2&=(1)^2
\\
\left(\sqrt{2x+1}\right)^2-2(\sqrt{2x+1})(\sqrt{x})+\left(\sqrt{x}\right)^2&=(1)^2
&(\text{use }(a+b)^2=a^2+2ab+b^2)
\\
2x+1-2\sqrt{x(2x+1)}+x&=1
\\
3x+1-2\sqrt{x(2x+1)}&=1
.\end{align*}
Using the properties of equality, the equation above is equivalent to
\begin{align*}
3x+1-1&=2\sqrt{x(2x+1)}
\\
3x&=2\sqrt{x(2x+1)}
.\end{align*}
Squaring both sides again, the equation above is equivalent to
\begin{align*}
(3x)^2&=\left(2\sqrt{x(2x+1)}\right)^2
\\
9x^2&=4[x(2x+1)]
\\
9x^2&=8x^2+4x
.\end{align*}
Using the properties of equality, the equation above is equivalent to
\begin{align*}
9x^2-8x^2-4x&=0
\\
x^2-4x&=0
\\
x(x-4)&=0
&(\text{factor }x)
.\end{align*}
Equating each factor to zero (Zero Product Property), then
\begin{array}{l|r}
x=0 & x-4=0
\\
& x=4
.\end{array}
Since both sides of the original equation were raised to an even power, then checking of solutions is a must. That is,
\begin{array}{l|r}
\text{If }x=0: & \text{If }x=4
\\\\
\sqrt{2(0)+1}-\sqrt{0}\overset{?}=1 & \sqrt{2(4)+1}-\sqrt{4}\overset{?}=1
\\
\sqrt{1}-0\overset{?}=1 & \sqrt{9}-2\overset{?}=1
\\
1-0\overset{?}=1 & 3-2\overset{?}=1
\\
1\overset{\checkmark}=1 & 1\overset{\checkmark}=1
.\end{array}
Hence, the solution set of the equation $
\sqrt{2x+1}-\sqrt{x}=1
$ is $
\{0,4\}
$.
