Help on solving the equation $\frac{\sqrt{a+x}}{\sqrt{a}+\sqrt{a+x}}=\frac{\sqrt{a-x}}{\sqrt{a}-\sqrt{a-x}}$

Question

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!

Answer 1

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