Rationalise the denominator of the following

Solution:

(i) Let $E=\frac{2}{3 \sqrt{3}}$

For rationalising the denominator, multiplying numerator and denominator by $\sqrt{3}$,

$E=\frac{2}{3 \sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$

$=\frac{2 \sqrt{3}}{3 \times 3}=\frac{2 \sqrt{3}}{9}$

(ii) Let $E=\frac{\sqrt{40}}{\sqrt{3}}$

For rationalising the denominator, multiplying 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 rationalising the denominator, multiplying 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, multiplying numerator and denominator by $\sqrt{41}+5$,

$E=\frac{16}{\sqrt{41}-5} \times \frac{\sqrt{41}+5}{\sqrt{41}+5}$

$=\frac{16(\sqrt{41}+5)}{(\sqrt{41})^{2}-(5)^{2}} \quad$ [using identity, $\left.(a-b)(a+b)=a^{2}-b^{2}\right]$

$=\frac{16(\sqrt{41}+5)}{41-25}$

$=\frac{16(\sqrt{41}+5)}{16}=\sqrt{41}+5$

(v) Let $E=\frac{2+\sqrt{3}}{2-\sqrt{3}}$

For rationalising the denominator, multiplying 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, $(a+b)^{2}=a^{2}+2 a b+b^{2}$ ]

$=4+3+4 \sqrt{3}=7+4 \sqrt{3}$

(vi) Let $E=\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}$

For rationalising the denominator, multiplying 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, $(a-b)(a+b)=a^{2}-b^{2}$ ]

$=\frac{\sqrt{6}(\sqrt{2}-\sqrt{3})}{2-3}=\frac{\sqrt{6}(\sqrt{2}-\sqrt{3})}{-1}=\sqrt{6}(\sqrt{3}-\sqrt{2})$

$=\sqrt{18}-\sqrt{12}=\sqrt{9 \times 2}-\sqrt{4 \times 3}=3 \sqrt{2}-2 \sqrt{3}$

(vii) Let $E=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$

For rationalising the denominator, multiplying 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, $(a-b)(a+b)=a^{2}-b^{2}$ ]

$=\frac{(\sqrt{3})^{2}+(\sqrt{2})^{2}+2 \sqrt{3} \sqrt{2}}{3-2}$

[using identity, $(a+b)^{2}=a^{2}+b^{2}+2 a b$ ]

$=3+2+2 \sqrt{6}=5+2 \sqrt{6}$

(viii) Let $E=\frac{3 \sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$

For rationalising the denominator, multiplying 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, multiplying 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, $\left.(a+b)(a-b)=a^{2}-b^{2}\right]$

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

$=\frac{18+8 \sqrt{6}}{30}=\frac{9+4 \sqrt{6}}{15}$