Question

Rationalise the denominator of the following:
$\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}$

Answer 1

$\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}$

multiplying both numerator and denominator with $\sqrt{2}-\sqrt{3}$

$=\frac{\left(\sqrt{6}\right)(\sqrt{2}-\sqrt{3})}{\sqrt{2}+\sqrt{3}(\sqrt{2}-\sqrt{3})}$

$=\frac{\sqrt{6}\sqrt{2}-\sqrt{6}\sqrt{3}}{\sqrt{2}\sqrt{2}-\sqrt{3}\sqrt{3}}$

$=\frac{\sqrt{6}\sqrt{2}-\sqrt{6}\sqrt{3}}{2-3}$

$=\frac{\sqrt{12}-\sqrt{18}}{2-3}$

$=\sqrt{18}-\sqrt{12}$