Question

Rationalise the denominator of the following:
$\frac{4\sqrt{3}+5\sqrt{2}}{\sqrt{48}+\sqrt{18}}$

Answer 1

$\frac{4\sqrt{3}+5\sqrt{2}}{\sqrt{48}+\sqrt{18}}$

Dividing numerator and denominator by $\sqrt{48}-\sqrt{18}$

$=\frac{4\sqrt{3}+5\sqrt{2}(\sqrt{48}-\sqrt{18})}{\sqrt{48}+\sqrt{18}(\sqrt{48}-\sqrt{18})}$

$=\frac{4\sqrt{3}\sqrt{48}-4\sqrt{3}\sqrt{18}+5\sqrt{2}\sqrt{48}-5\sqrt{2}\sqrt{18}}{\sqrt{48}\sqrt{48}-\sqrt{18}\sqrt{18}}$

$=\frac{4\times 4\sqrt{3}\sqrt{3}-4\times 3\sqrt{3}\sqrt{2}+5\times 4\sqrt{2}\sqrt{3}-5\times 3\sqrt{2}\sqrt{2}}{48-18}$

$=\frac{4\times 4\times 3-4\times 3\sqrt{6}+5\times 4\sqrt{6}-5\times 3\times 2}{30}$

$=\frac{48-12\sqrt{6}+20\sqrt{6}-30}{30}$

$=\frac{18+8\sqrt{6}}{30}$

$=\frac{9+4\sqrt{6}}{15}$