Question

Solve the following equation:
$\frac{2+x}{\sqrt{2}+\sqrt{2+x}}+\frac{2-x}{\sqrt{2}-\sqrt{2+x}}=2\sqrt{2}$

Answer 1

$\frac{2+x}{\sqrt{2}+\sqrt{2+x}}+\frac{2-x}{\sqrt{2}-\sqrt{2+x}}=2\sqrt{2}⟹\sqrt{2}+\sqrt{2+x}\ne 0⟹x\ne 0$

$(2+x)(\sqrt{2}-\sqrt{2+x})+(2-x)(\sqrt{2}+\sqrt{2+x})=2\sqrt{2}\left(\right(\sqrt{2}{)}^2-\left(\sqrt{2+x}{)}^2\right)$

$⟹2\sqrt{2}-x\sqrt{2+x}=-\sqrt{2}x$

$⟹\sqrt{2}(2+x)=x\sqrt{2+x}$

$⟹2+x=0⟹x=-2$ or $\sqrt{4+2x}=x⟹x^2-2x-4=0⟹x=\frac{2\pm \sqrt{20}}{2}=1\pm \sqrt{5}$