Question

Question 10
Rationalize the denominator of the following
$\left(i\right)\frac{2}{3\sqrt{3}}\left(ii\right)\frac{\sqrt{40}}{\sqrt{3}}\left(iii\right)\frac{3+\sqrt{2}}{4\sqrt{2}}\left(iv\right)\frac{16}{\sqrt{41-5}}\left(v\right)\frac{2+\sqrt{3}}{2-\sqrt{3}}\left(vi\right)\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}\left(vii\right)\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$
$\left(viii\right)\frac{3\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$
$\left(ix\right)\frac{4\sqrt{3}+5\sqrt{2}}{\sqrt{48}+\sqrt{18}}$
Thinking process
For rationalizing the denominator multiplying numerator by the conjugate of the denominator term and simplify it to get the result.

Answer 1

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

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

(iii)
Let $E=\frac{3+\sqrt{2}}{4\sqrt{2}}$
For rationailsing the denominator, multiply numerator and denominator by $\sqrt{2}$,
$E=\frac{3+\sqrt{2}}{4\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}=\frac{3\sqrt{2}+(\sqrt{2}{)}^2}{4(\sqrt{2}{)}^2}=\frac{3\sqrt{2}+2}{4\times 2}=\frac{3\sqrt{2}+2}{8}$

(iv)
Let $E=\frac{16}{\sqrt{41}-5}$
For rationalising the denominator, multiply numerator and denominator by $\sqrt{41}+5$
$E=\frac{16}{\sqrt{41}-5}\times \frac{\sqrt{41}+5}{\sqrt{41}+5}$
$=\frac{16(\sqrt{41}+5)}{(\sqrt{41}{)}^2-(5{)}^2}$ [Using identity, $(a-b)(a+b)=a^2-b^2$]
$=\frac{16(\sqrt{41}+5)}{16}=\sqrt{41}+5$

(v)
Let
$E=\frac{2+\sqrt{3}}{2-\sqrt{3}}$
For rationailsing the denominator, multiply numerator and denominator by $2+\sqrt{3}$
$E=\frac{2+\sqrt{3}}{2-\sqrt{3}}\times \frac{2+\sqrt{3}}{2+\sqrt{3}}=\frac{(2+\sqrt{3}{)}^2}{(2{)}^2-(\sqrt{3}{)}^2}$
[Using identity, $(a-b)(a+b)=a^2-b^2]$
$=\frac{2^2+(\sqrt{3}{)}^2+2\times 2\times \sqrt{3}}{4-3}$
[Using identity $(a+b{)}^2=a^2+2ab+b^2]$
$=4+3+4\sqrt{3}=7+4\sqrt{3}$

(vi)
Let $E=\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}$
For rationalising the denominator, multiply numerator and denominator by $\sqrt{2}-\sqrt{3}$
$E=\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}\times \frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}}=\frac{\sqrt{6}(\sqrt{2}-\sqrt{3})}{(\sqrt{2}{)}^2-(\sqrt{3}{)}^2}$
[Using identity, $(a-b)(a+b)=a^2-b^2]$
$\frac{\sqrt{6}(\sqrt{2}-\sqrt{3})}{(\sqrt{2}{)}^2-(\sqrt{3}{)}^2}=\frac{\sqrt{6}(\sqrt{2}-\sqrt{3})}{-1}=\sqrt{6}(\sqrt{3}-\sqrt{2})=\sqrt{18}-\sqrt{12}=\sqrt{9\times 2}-\sqrt{4\times 3}=3\sqrt{2}-2\sqrt{3}$

(vii)
For rationalising the denominator, multiply numerator and denominator by $\sqrt{3}+\sqrt{2}$
$E=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}-\frac{(\sqrt{3}+\sqrt{2}{)}^2}{(\sqrt{3}{)}^2-(\sqrt{2}{)}^2}$
[Using identity, $(a-b)(a+b)=a^2-b^2]$
$=3+2+2\sqrt{6}=5+2\sqrt{6}$

(viii)
Let $E=\frac{3\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$
For rationalising the denominator, multiply numerator and denominator by $\sqrt{5}+\sqrt{3}$
$E=\frac{3\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}\times \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}=\frac{3\sqrt{5}(\sqrt{5}+\sqrt{3})+\sqrt{3}(\sqrt{5}+\sqrt{3})}{(\sqrt{5}{)}^2-(\sqrt{3}{)}^2}$ [Using identity, $(a+b)(a-b)=a^2-b^2]$
$=\frac{15+3\sqrt{15}+\sqrt{15}+3}{5-3}=\frac{18+4\sqrt{15}}{2}=9+2\sqrt{15}$

(ix)
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 $4\sqrt{3}-3\sqrt{2}$,
$=\frac{4\sqrt{3}+5\sqrt{2}}{4\sqrt{3}+3\sqrt{2}}\times \frac{(4\sqrt{3}-3\sqrt{2})}{(4\sqrt{3}-3\sqrt{2})}=\frac{4\sqrt{3}(4\sqrt{3}-3\sqrt{2})+5\sqrt{2}(4\sqrt{3}-3\sqrt{2})}{(4\sqrt{3}{)}^2-(3\sqrt{2}{)}^2}$
[Using identity, $(a+b)(a-b)=a^2-b^2]$
$=\frac{48-12\sqrt{6}+20\sqrt{6}-30}{30}=\frac{18+8\sqrt{6}}{30}=\frac{9+4\sqrt{6}}{15}$