Answer
$\{ 5\}$
Work Step by Step
Start by squaring both sides of the equation to get rid of the radical sign. Expand the right side carefully as shown.
\begin{equation}
\begin{aligned}
\sqrt{x-4}+\sqrt{x+4}&=4\\
\left(\sqrt{x+4}\right)^2&= \left(4-\sqrt{x-4}\right)^2\\\
x+4& =16-8\sqrt{x-4}+x-4 \\
x+4& =12-8\sqrt{x-4}+x \\
x-x+4-12& =-8\sqrt{x-4}\\
-8& = -8\sqrt{x-4}\\
1& = \sqrt{x-4}\\
1^2&=\left(\sqrt{x-4}\right)^2\\
1&= x-4\\
1+4&= x\\
5&= x
\end{aligned}
\end{equation}
In line 3, I have used the fact that $((a-b)^2= a^2-2ab+b^2$ to expand $\left(4-\sqrt{x-4}\right)^2= 16-8\sqrt{x-4}+x-4 $.
Check:
\begin{equation}
\begin{aligned}
\sqrt{5-4}+\sqrt{5+4}&=4\\
\sqrt{1}+\sqrt{9}&=4\\
1+3&=4\\
4&= 4
\end{aligned}
\end{equation}
