Rationalise the denominator in each of the following

Solution:

(i) Let $E=\frac{4}{\sqrt{3}}$

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

$E=\frac{4}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{4 \sqrt{3}}{3}$

$=\frac{4}{3} \times 1.732=\frac{6.928}{3}=2.309$          [put $\sqrt{3}=1.732$ ]

(ii) Let $E=\frac{6}{\sqrt{6}}$

For rationalising the denominator, multiplying numerator and denominator by $\sqrt{6}$, we get

$E=\frac{6}{\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}}=\frac{6 \sqrt{6}}{6}=\sqrt{2} \times \sqrt{3}$

$=1.414 \times 1.732=2.449$             [put $\sqrt{2}=1.414$ and $\sqrt{3}=1.732$ ]

(iii) Let $E=\frac{\sqrt{10}-\sqrt{5}}{2}=\frac{\sqrt{5} \sqrt{2}-\sqrt{5}}{2}=\frac{\sqrt{5}(\sqrt{2}-1)}{2}$       $[\because \sqrt{10}=\sqrt{2} \cdot \sqrt{5}]$

$=\frac{2.236(1.414-1)}{2}=1.118 \times 0.414=0.46285 \cong 0.463$

(iv) Let $E=\frac{\sqrt{2}}{2+\sqrt{2}}$

For rationalising the denominator, multiplying numerator and denominator by $2-\sqrt{2}$, we get

$=\frac{\sqrt{2}}{2+\sqrt{2}} \times \frac{2-\sqrt{2}}{2-\sqrt{2}}=\frac{\sqrt{2}(2-\sqrt{2})}{(2)^{2}-(\sqrt{2})^{2}}$            [using identity, $(a-b)(a+b)=a^{2}-b^{2}$ ]

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

$=\sqrt{2}-1=1.414-1=0.414$           [put $\sqrt{2}=1.414$ ]

(v) Let $E=\frac{1}{\sqrt{3}+\sqrt{2}}$

For rationalising the denominator multiplying numerator and denominator by $\sqrt{3}-\sqrt{2}$, we get

$\frac{1}{\sqrt{3}+\sqrt{2}} \times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}=\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3})^{2}-(\sqrt{2})^{2}}$

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

$=\frac{\sqrt{3}-\sqrt{2}}{3-2}=\sqrt{3}-\sqrt{2}$

$=1.732-1.414=0.318$       [put $\sqrt{3}=1.732$ and $\sqrt{2}=1.414$ ]