Question

If $\frac{\sqrt{2}}{\sqrt{6}-\sqrt{2}}-\frac{\sqrt{3}}{\sqrt{6}+\sqrt{2}}$ is equal to $\frac{2\sqrt{3}+2-3\sqrt{2}+\sqrt{6}}{m}$ then value of m is

Answer 1

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