Question

Solve $\frac{\sqrt{2-x}+\sqrt{2+x}}{\sqrt{2-x}-\sqrt{2+x}}=3$

Answer 1

$$\begin{gathered} Wehave: \\ \frac{\sqrt{2-x}+\sqrt{2+x}}{\sqrt{2-x}-\sqrt{2+x}}=3 \\ \Rightarrow \frac{\sqrt{2-x}+\sqrt{2+x}+\sqrt{2-x}-\sqrt{2+x}}{\sqrt{2-x}+\sqrt{2+x}-\sqrt{2-x}+\sqrt{2+x}}=\frac{3+1}{3-1}\left(\mathrm{Componendo}\mathrm{and}\mathrm{dividendo}\right) \\ \Rightarrow \frac{2\sqrt{2-x}}{2\sqrt{2+x}}=\frac{4}{2} \\ \Rightarrow \sqrt{2-x}=2\sqrt{2+x} \\ \mathrm{Squaring}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}: \\ \left(2-x\right)=4\left(2+x\right) \\ \Rightarrow 2-x=8+4x \\ \Rightarrow -5x=6 \\ \Rightarrow x=\frac{-6}{5} \end{gathered}$$