Solve $\sqrt{x-4} + 10 = \sqrt{x+4}$ Solve: $$\sqrt{x-4} + 10 = \sqrt{x+4}$$
Little help here? >.<

We will assume that $x$ ranges over the reals $\ge 4$, to make sure that the square roots are real. Note that
$$\sqrt{x+4}-\sqrt{x-4}=\frac{(\sqrt{x+4}-\sqrt{x-4})(\sqrt{x+4}+\sqrt{x-4})}{\sqrt{x+4}+\sqrt{x-4}} =\frac{8}{\sqrt{x+4}+\sqrt{x-4}} .$$
For $x\ge 4$, $\sqrt{x+4}+\sqrt{x-4}\ge 2\sqrt{2}$. It follows that $\sqrt{x+4}-\sqrt{x-4}\le \dfrac{8}{2\sqrt{2}}=2\sqrt{2}$ for all $x\ge 4$. In particular, $\sqrt{x+4}-\sqrt{x-4}$ cannot be equal to $10$.
