Radical equation help

Question

Find all real values of x for which $$\frac{1}{\sqrt{x}+\sqrt{x-2}}+\frac{1}{\sqrt{x}+\sqrt{x+2}}=\frac{1}{4}$$ I tried expanding and simplifing terms, but everything becomes too complicated, and the question is meant to be done under 15 minutes. Is there a quick way of solving this?

Answer 1

multiplying the denominators and numerators with $$\sqrt{x}-\sqrt{x-2}$$ and $$\sqrt{x}-\sqrt{x+2}$$ we get $$\sqrt{x}-\sqrt{x-2}-\sqrt{x}+\sqrt{x+2}=\frac{1}{2}$$ can you proceed? now we write $$\sqrt{x+2}=1/2+\sqrt{x-2}$$ after squaring and simplifying we obtain $$\sqrt{x-2}=\frac{15}{4}$$ thus we get the solution $$x=2+\left(\frac{15}{4}\right)^2$$