Question

Rationalise the denominator and simplify
$\frac{\sqrt{48}+\sqrt{32}}{\sqrt{27}-\sqrt{18}}$

Answer 1

$\frac{\sqrt{48}+\sqrt{32}}{\sqrt{27}-\sqrt{18}}$

$=\frac{\sqrt{48}+\sqrt{32}}{\sqrt{27}-\sqrt{18}}$$\times$$\frac{\sqrt{27}+\sqrt{18}}{\sqrt{27}+\sqrt{18}}$

$=\frac{(\sqrt{48}+\sqrt{32})(\sqrt{27}+\sqrt{18})}{27-18}$ $[\because a^2-b^2=(a-b\left)\right(a+b\left)\right]$

$=\frac{(48\times 27+48\times 18+32\times 27+32\times 18)}{27-18}$

$=\frac{(1296+864+864+576)}{9}$

$=\frac{3600}{9}$

$=400$