Question

Simplify:

If $\frac{\sqrt{2}}{\sqrt{6}-\sqrt{2}}-\frac{\sqrt{3}}{\sqrt{6}+\sqrt{2}}$ is $\frac{2\sqrt{3}+2-3\sqrt{2}+\sqrt{m}}{4}$, then $m$ is

Answer 1

We rationalize the denominators with the respective factors,
We get, $\frac{\sqrt{2}}{\sqrt{6}-\sqrt{2}}\times \frac{\sqrt{6}+\sqrt{2}}{\sqrt{6}+\sqrt{2}}-\frac{\sqrt{3}}{\sqrt{6}+\sqrt{2}}\times \frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}-\sqrt{2}}$
$=\frac{\sqrt{12}+2+\sqrt{6}-\sqrt{18}}{6-2}$
$=\frac{2\sqrt{3}+2+\sqrt{6}-3\sqrt{2}}{4}$