
Originally Posted by
Soroban
Hello, hc0company!
Square both sides: .$\displaystyle \left(\sqrt{x+a^2} - \sqrt{x-2a^2}\right)^2 \;=\;\left(\sqrt{2x-5a^2}\right)^2$
. . $\displaystyle x + a^2 - 2\sqrt{(x+a^2)(x-2a^2)} + x - 2a^2 \;=\;2x-5a^2$
. . $\displaystyle 2\sqrt{(x+a^2)(x - 2a^2)} \;=\;4a^2 \quad\Rightarrow\quad \sqrt{(x+a^2)(x-2a^2)} \;=\;2a^2 $
Square both sides: .$\displaystyle \left(\sqrt{(x+a^2)(x-2a^2)}\right)^2 \;=\;\left(2a^2\right)^2$
. . $\displaystyle (x+a^2)(x-2a^2) \;=\;4a^4 \quad\Rightarrow\quad x^2 - a^2x - 6a^4 \;=\;0$
Factor: .$\displaystyle (x - 3a^2)(x + 2a^2) \;=\;0$
Therefore: .$\displaystyle \begin{array}{cccc}x - 3a^2 \:=\:0 & \Rightarrow & \boxed{x \:=\:3a^2} \\ \\[-4mm]
x + 2a^2 \:=\: 0 & \Rightarrow & {\color{red}\rlap{/////////}}x \:=\:-2a^2 & \text{extraneous root} \end{array}$