Answer
$x=4$
Work Step by Step
Squaring both sides of the given equation, $
\sqrt{x}=1+\sqrt{2x-7}
,$ results to
\begin{align*}
\left(\sqrt{x}\right)^2&=\left(1+\sqrt{2x-7}\right)^2
\\
x&=(1)^2+2(1)\left(\sqrt{2x-7}\right)+\left(\sqrt{2x-7}\right)^2
&\left(\text{Use }(a+b)^2=a^2+2ab+b^2\right)
\\
x&=1+2\sqrt{2x-7}+2x-7
\\
x-2x-1+7&=2\sqrt{2x-7}
\\
-x+6&=2\sqrt{2x-7}
.\end{align*}
Squaring both sides of the equation results to
\begin{align*}
(-x+6)^2&=\left(2\sqrt{2x-7}\right)
\\
(x)^2+2(-x)(6)+(6)^2&=4(2x-7)
&\left(\text{Use }(a+b)^2=a^2+2ab+b^2\right)
\\
x^2-12x+36&=4(2x)+4(-7)
&\left(\text{Use Distributive Property}\right)
\\
x^2-12x+36&=8x-28
\\
x^2-12x+36-8x+28&=0
\\
x^2+(-12x-8x)+(36+28)&=0
&\left(\text{Combine like terms}\right)
\\
x^2-20x+64&=0
.\end{align*}
Using the factoring of quadratic trinomials, the factored form of the equation above is
\begin{align*}
(x-16)(x-4)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving for the variable, then
\begin{align*}
x-16&=0
\\
x&=16
\\\\\text{OR}\\\\
x-4&=0
\\
x&=4
.\end{align*}
Check the solution: $
\sqrt{x}=1+\sqrt{2x-7}
$. That is,
\begin{align*}
\text{If }x=4:\\
\sqrt{4}&\overset{?}=1+\sqrt{2(4)-7}
\\
2&\overset{?}=1+\sqrt{8-7}
\\
2&\overset{?}=1+\sqrt{1}
\\
2&\overset{?}=1+1
\\
2&\overset{\checkmark}=2
\text{ (Hence, $x=4$ is a solution)}
\\\\
\text{If }x=16:\\
\sqrt{16}&\overset{?}=1+\sqrt{2(16)-7}
\\
4&\overset{?}=1+\sqrt{32-7}
\\
4&\overset{?}=1+\sqrt{25}
\\
4&\overset{?}=1+5
\\
4&\ne6
\text{ (Hence, $x=16$ is NOT a solution)}
.\end{align*}
Hence, the only solution to the equation $
\sqrt{x}=1+\sqrt{2x-7}
$ is $
x=4
.$
