Help on solving the equation $\frac{\sqrt{a+x}}{\sqrt{a}+\sqrt{a+x}}=\frac{\sqrt{a-x}}{\sqrt{a}-\sqrt{a-x}}$ Could you give me some help on finding the roots (if any) of the following equation:
$$
\frac{\sqrt{a+x}}{\sqrt{a}+\sqrt{a+x}}=\frac{\sqrt{a-x}}{\sqrt{a}-\sqrt{a-x}}
$$
I tried to apply some classic approaches, but I had no luck... Could you lend me a hand? Thanks in advance!

$$\frac{\sqrt{a+x}}{\sqrt{a}+\sqrt{a+x}}=\frac{\sqrt{a-x}}{\sqrt{a}-\sqrt{a-x}}$$
$$\frac{\sqrt{1+\frac xa}}{\sqrt{1}+\sqrt{1+\frac xa}}=\frac{\sqrt{1-\frac xa}}{\sqrt{1}-\sqrt{1-\frac xa}}$$
$$\frac xa \mapsto y$$
$$\frac{\sqrt{1+y}}{1+\sqrt{1+y}}=\frac{\sqrt{1-y}}{1-\sqrt{1-y}}$$
$$\sqrt{1+y}-\sqrt{1-y^2}=\sqrt{1-y}+\sqrt{1-y^2}$$
$$\sqrt{1+y}-\sqrt{1-y}=2\sqrt{1-y^2}$$
$$1+y+1-y-2\sqrt{1-y^2}=4(1-y^2)$$
$$4y^2-2=2\sqrt{1-y^2}$$
$$2y^2-1=\sqrt{1-y^2}$$
$$4y^4-4y^2+1=1-y^2$$
$$4y^4-3y^2=0$$
$$y^2(4y^2-3)=0$$
$$y=0,\quad y=\pm\frac{\sqrt3}2$$
$$x=0,\quad x=\pm\frac{\sqrt3}2a$$
