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Mathematics

Rationalise the denominator of the following :

(i)1022+3(ii)73−5248+18(iii)13−2+1\begin{matrix} \text{(i)} & \dfrac{10}{2\sqrt{2} + \sqrt{3}} \\[1.5em] \text{(ii)} & \dfrac{7\sqrt{3} - 5\sqrt{2}}{\sqrt{48} + \sqrt{18}} \\[1.5em] \text{(iii)} & \dfrac{1}{\sqrt{3} - \sqrt{2} + 1} \\[1.5em] \end{matrix}

Rational Irrational Nos

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Answer

(i)\text{(i)}

1022+3\dfrac{10}{2\sqrt{2} + \sqrt{3}}

Let us rationalise the denominator,

Then,

1022+3=1022+3×22−322−3=10(22−3)(22)2−(3)2=10×22−10×3(22)2−(3)2=10(22−3)8−3=10(22−3)5=2(22−3)\dfrac{10}{2\sqrt{2} + \sqrt{3}} = \dfrac{10}{2\sqrt{2} + \sqrt{3}} × \dfrac{2\sqrt{2} - \sqrt{3}}{2\sqrt{2} - \sqrt{3}} \\[1.5em] = \dfrac{10({2\sqrt{2} - \sqrt{3}})}{(2\sqrt{2})^2 - (\sqrt{3})^2} \\[1.5em] = \dfrac{10 × 2\sqrt{2} - 10 × \sqrt{3}}{(2\sqrt{2})^2 - (\sqrt{3})^2} \\[1.5em] = \dfrac{10(2\sqrt{2} - \sqrt{3})}{8 - 3} \\[1.5em] = \dfrac{10(2\sqrt{2} - \sqrt{3})}{5} \\[1.5em] \bold{= 2(2\sqrt{2} - \sqrt{3}) } \\[1.5em]

(ii)\text{(ii)} Since, it is given that

73−5248+18\dfrac{7\sqrt{3} - 5\sqrt{2}}{\sqrt{48} + \sqrt{18}}

Let us rationalise the denominator,

73−5248+18×48−1848−18=73×48−73×18−52×48+52×18(48)2−(18)2=7144−754−596+53648−18=7×12−72×3×3×3−52×2×2×2×2×3+52×2×3×330=84−216−206+3030=84−216−206+3030=114−41630=11430−41630=5715−41630\dfrac{7\sqrt{3} - 5\sqrt{2}}{\sqrt{48} + \sqrt{18}} × \dfrac{\sqrt{48} - \sqrt{18}}{\sqrt{48} - \sqrt{18}} \\[1.5em] = \dfrac{7\sqrt{3} × \sqrt{48} - 7\sqrt{3} × \sqrt{18} - 5\sqrt{2} × \sqrt{48} + 5\sqrt{2} × \sqrt{18} }{(\sqrt{48})^2 -(\sqrt{18})^2} \\[1.5em] = \dfrac{7\sqrt{144} - 7\sqrt{54} - 5\sqrt{96} +5\sqrt{36}}{48 - 18} \\[1.5em] = \dfrac{7 × 12 - 7\sqrt{2 × 3 × 3 × 3 } - 5\sqrt{2 × 2 × 2 × 2 × 2 × 3} +5\sqrt{2 × 2 × 3 × 3}}{30} \\[1.5em] = \dfrac{84 - 21\sqrt{6} - 20\sqrt{6} + 30}{30} \\[1.5em] = \dfrac{84 - 21\sqrt{6} - 20\sqrt{6} + 30}{30} \\[1.5em] = \dfrac{114 - 41{\sqrt6}}{30} \\[1.5em] = \dfrac{114}{30} - \dfrac{41\sqrt{6}}{30} \\[1.5em] = \bold{\dfrac{57}{15} - \dfrac{41\sqrt{6}}{30}} \\[1.5em]

(iii)\text{(iii)}

13−2+1\dfrac{1}{\sqrt{3} - \sqrt{2} + 1}

Let us rationalise the denominator,

Then,

13−2+1=13−(2−1)×3+(2−1)3+(2−1)=3+2−1(3)2−(2−1)2=3+2−1(3)2−((2)2−2×2+1)=3+2−13−(2−22+1)=3+2−13−2+22−1=3+2−122=3+2−122×22=2(3+2−1)22×2=2×3+2×2−222×2=6+2−24=2+6−24\dfrac{1}{\sqrt{3} - \sqrt{2} + 1} = \dfrac{1}{\sqrt{3} - (\sqrt{2} - 1)} × \dfrac{\sqrt{3} + (\sqrt{2} - 1)}{\sqrt{3} + (\sqrt{2} - 1)} \\[1.5em] =\dfrac{{\sqrt{3} + \sqrt{2}} - 1}{{(\sqrt{3})^2 }- (\sqrt{2} - 1)^2} \\[1.5em] =\dfrac{\sqrt{3} + \sqrt{2} - 1}{(\sqrt{3})^2 - ( (\sqrt{2})^2 - 2 × \sqrt{2} + 1)} \\[1.5em] =\dfrac{\sqrt{3} + \sqrt{2} - 1}{3 - (2 - 2\sqrt{2} + 1)} \\[1.5em] =\dfrac{\sqrt{3} + \sqrt{2} - 1}{3 - 2 + 2\sqrt{2} -1} \\[1.5em] =\dfrac{\sqrt{3} + \sqrt{2} - 1}{2\sqrt{2}} \\[1.5em] =\dfrac{\sqrt{3} + \sqrt{2} - 1}{2\sqrt{2}} × \dfrac{\sqrt{2}}{\sqrt{2}} \\[1.5em] =\dfrac{\sqrt{2}(\sqrt{3} + \sqrt{2} - 1)}{2\sqrt{2} × \sqrt{2}} \\[1.5em] =\dfrac{\sqrt{2} × \sqrt{3} + \sqrt{2} × \sqrt{2} - \sqrt{2}}{2\sqrt{2} × \sqrt{2}} \\[1.5em] = \dfrac{\sqrt{6} + 2 -\sqrt{2}}{4} \\[1.5em] \bold{{=} \dfrac{2 + \sqrt{6} -\sqrt{2}}{4}} \\[1.5em]

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