Answer
$x=5$
Work Step by Step
Squaring both sides of the given equation, $
\sqrt{2x-1}-4=-\sqrt{x-4}
,$ results to
\begin{array}{l}\require{cancel}
(\sqrt{2x-1}-4)^2=(-\sqrt{x-4})^2
\\
(\sqrt{2x-1})^2-2(\sqrt{2x-1})(4)+(4)^2=x-4
\\
2x-1-8\sqrt{2x-1}+16=x-4
\\
(2x-x)+(-1+16+4)=8\sqrt{2x-1}
\\
x+19=8\sqrt{2x-1}
.\end{array}
Squaring both sides for the second time results to
\begin{array}{l}\require{cancel}
(x+19)^2=(8\sqrt{2x-1})^2
\\
(x)^2+2(x)(19)+(19)^2=64(2x-1)
\\
x^2+38x+361=128x-64
\\
x^2+(38x-128x)+(361+64)=0
\\
x^2-90x+425=0
\\
(x-85)(x-5)=0
.\end{array}
Equating each factor to zero (Zero Product Property) and then solving for the variable, the solutions are $
x=\{ 5,85 \}
.$
If $x=-85,$ the original equation becomes
\begin{array}{l}\require{cancel}
\sqrt{2x-1}-4=-\sqrt{x-4}
\\
\sqrt{2(-85)-1}-4=-\sqrt{-85-4}
\\
\sqrt{-170-1}-4=-\sqrt{-89}
\\
\sqrt{-171}-4=-\sqrt{-89}
.\end{array}
Since the radicands are negative and the indices are even, then the radicals result to nonreal numbers. Hence, only $
x=5
$ is the solution.
