Solve for x Given Two Square Roots: Algebra Problem

Question

I am trying to solve for all real numbers for $x$ given $5=\sqrt{9-x^2}+\sqrt{16-x^2}$.

The answer is that $x=\pm \frac{12}{5}$. I am looking for a clean way to do it. I am stumped, and I feel like I am missing something about conjugates here.

Answer 1

There is no clean way that I know of. We have

$5=\sqrt{9-x^2}+\sqrt{16-x^2} \implies $

$5-\sqrt{9-x^2}=\sqrt{16-x^2} \implies $

$(5-\sqrt{9-x^2})^2=16-x^2 \implies $

$25+9-x^2-10\sqrt{9-x^2} = 16-x^2 \implies $

$34-x^2-10\sqrt{9-x^2} = 16-x^2 \implies $

$18 =10\sqrt{9-x^2} \implies $

$\frac{9}{5} = \sqrt{9-x^2} \implies $

$\frac{81}{25} = 9-x^2 \implies $

$ \frac{-144}{25} = -x^2 \implies $

$ x = \pm \frac{12}{5} $