Answer
$x=8$
Work Step by Step
Isolating the radical and then squaring both sides of the given equation, $
x-\sqrt{2x}=4
,$ then,
\begin{array}{l}\require{cancel}
x-4=\sqrt{2x}
\\\\
\left( x-4 \right)^2=\left( \sqrt{2x} \right)^2
\\\\
(x)^2+2(x)(-4)+(-4)^2=2x
\\\\
x^2-8x+16=2x
\\\\
x^2+(-8x-2x)+16=0
\\\\
x^2-10x+16=0
.\end{array}
Using $\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ (or the Quadratic Formula), the solutions of the quadratic equation, $
x^2-10x+16=0
,$ are
\begin{array}{l}\require{cancel}
\dfrac{-(-10)\pm\sqrt{(-10)^2-4(1)(16)}}{2(1)}
\\\\=
\dfrac{10\pm\sqrt{100-64}}{2}
\\\\=
\dfrac{10\pm\sqrt{36}}{2}
\\\\=
\dfrac{10\pm6}{2}
\\\\=
\dfrac{10-6}{2}
\text{ OR }
\dfrac{10+6}{2}
\\\\=
\dfrac{4}{2}
\text{ OR }
\dfrac{16}{2}
\\\\=
2
\text{ OR }
8
.\end{array}
Upon checking, only $
x=8
$ satisfies the original equation.