Question

Rationalize the denominator of the following
$\left(i\right)\frac{2}{3\sqrt{3}}\left(ii\right)\frac{\sqrt{40}}{\sqrt{3}}\left(iii\right)\frac{4\sqrt{3}+5\sqrt{2}}{\sqrt{48}+\sqrt{18}}$

Answer 1

$\text{(i) Let}E=\frac{2}{3\sqrt{3}}$
For rationalising the denominator, multiply
numerator and denominator by $\sqrt{3}$,
$E=\frac{2}{3\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=\frac{2\sqrt{3}}{3\times 3}=\frac{2\sqrt{3}}{9}$

$\text{(ii)Let}E=\frac{\sqrt{40}}{\sqrt{3}}$
For rationailsing the denominator, multiply numerator and denominator by $\sqrt{3}$,
$E=\frac{\sqrt{40}}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{40\times 3}}{(\sqrt{3}{)}^2}=\frac{\sqrt{120}}{3}=\frac{\sqrt{2\times 2\times 2\times 5\times 3}}{3}=\frac{2}{3}\sqrt{30}$

$\text{(iii) Let}E=\frac{4\sqrt{3}+5\sqrt{2}}{\sqrt{48}+\sqrt{18}}=\frac{4\sqrt{3}+5\sqrt{2}}{\sqrt{16\times 3}+\sqrt{9\times 2}}=\frac{4\sqrt{3}+5\sqrt{2}}{4\sqrt{3}+3\sqrt{2}}$
For rationalising the denominator, multiply numerator and denominator by $4\sqrt{3}-3\sqrt{2}$,
$=\frac{4\sqrt{3}+5\sqrt{2}}{4\sqrt{3}+3\sqrt{2}}\times \frac{(4\sqrt{3}-3\sqrt{2})}{(4\sqrt{3}-3\sqrt{2})}=\frac{4\sqrt{3}(4\sqrt{3}-3\sqrt{2})+5\sqrt{2}(4\sqrt{3}-3\sqrt{2})}{(4\sqrt{3}{)}^2-(3\sqrt{2}{)}^2}$
$\text{[Using identity,}(a+b)(a-b)=a^2-b^2]$
$=\frac{48-12\sqrt{6}+20\sqrt{6}-30}{30}=\frac{18+8\sqrt{6}}{30}=\frac{9+4\sqrt{6}}{15}$