Question

$$\sqrt{4a+x}-\sqrt{a-x}=\sqrt{a}$$

Solve for a (complex solution)

$\left\{\begin{matrix}a=-\frac{x}{3}\text{, }&\left(|\frac{arg(-x)}{2}-arg(-i\sqrt{x})|<\pi \text{ and }|-arg(-\frac{i\sqrt{x}}{3})+arg(-\sqrt{-x})|<\pi \right)\text{ or }\left(|\frac{arg(-x)}{2}-arg(i\sqrt{x})|<\pi \text{ and }|-arg(\frac{i\sqrt{x}}{3})+arg(-\sqrt{-x})|<\pi \right)\\a\neq 0\text{, }&x=0\end{matrix}\right.$

Solve for x (complex solution)

$\left\{\begin{matrix}\\x=0\text{, }&\text{unconditionally}\\x=-3a\text{, }&|-\frac{arg(a)}{2}+arg(\sqrt{x+4a}-\sqrt{a-x})|<\pi \end{matrix}\right.$

Solve for a

$a\geq 0$
$x=0$

Solve for x

$x=0$
$a\geq 0$