Question

Rationalise the denominator and simplify:

If $\frac{1}{2\sqrt{5}-\sqrt{3}}$ is $\frac{2\sqrt{m}+\sqrt{3}}{17}$, then $m$ is ?

Answer 1

We rationalize the denominator, $\frac{1}{2\sqrt{5}-\sqrt{3}}\ast \frac{2\sqrt{5}+\sqrt{3}}{2\sqrt{5}+\sqrt{3}}=\frac{2\sqrt{5}+\sqrt{3}}{4\ast 5-3}=\frac{2\sqrt{5}+\sqrt{3}}{17}$

$\therefore$ value of $m=5$