Answer
$x=\sqrt{a^2+36}$
Work Step by Step
$\sqrt{x+a}+\sqrt{x-a}=\sqrt{2}\sqrt{x+6}$ (Take the square)
$x+a+2\sqrt{(x+a)(x-a)}+x-a=2(x+6)$ (Simplify)
$2x+2\sqrt{x^2-a^2}=2x+12$ (Substract with $2x$)
$2\sqrt{x^2-a^2}=12$ (Divide by 2)
$\sqrt{x^2-a^2}=6$ (Take the square)
$x^2-a^2=36$
$x^2=a^2+36$ (Take the square root)
$x=\pm \sqrt{a^2+36}$
If $x=-\sqrt{a^2+36}$, then the value of $x+a<0$ and so it is not allowed because we have $\sqrt{x+a}$ in the original equation.
Meanwhile, if $x=\sqrt{a^2+36}$, then $x+a$, $x-a$, and $x+6$ are positive.
So, the only solution is $x=\sqrt{a^2+36}$.
