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By rationalising the denominator of each of the following; find, in each case, the value correct to two significant figures :

(i) 13−2\dfrac{1}{3 - \sqrt2}

(ii) 12+3\dfrac{1}{2 + \sqrt3}

(iii) 432−23\dfrac{4}{3\sqrt2 - 2\sqrt3}

Rational Irrational Nos

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Answer

(i) Since, the denominator = 3−23 - \sqrt2, its rationalizing factor = 3+23 + \sqrt2.

13−2=13−2×3+23+2=3+2(3)2−(2)2=3+29−2=3+27=3+1.417=4.417=0.63\dfrac{1}{3 - \sqrt2} = \dfrac{1}{3 - \sqrt2} \times \dfrac{3 + \sqrt2}{3 + \sqrt2}\\[1em] = \dfrac{3 + \sqrt2}{(3)^2 - (\sqrt2)^2}\\[1em] = \dfrac{3 + \sqrt2}{9 - 2}\\[1em] = \dfrac{3 + \sqrt2}{7}\\[1em] = \dfrac{3 + 1.41}{7}\\[1em] = \dfrac{4.41}{7}\\[1em] = 0.63

Hence, 13−2\dfrac{1}{3 - \sqrt2} = 0.63.

(ii) Since, the denominator = 2+32 + \sqrt3, its rationalizing factor = 2−32 - \sqrt3.

12+3=12+3×2−32−3=2−3(2)2−(3)2=2−34−3=2−31=2−3=2−1.73=0.27\dfrac{1}{2 + \sqrt3} = \dfrac{1}{2 + \sqrt3} \times \dfrac{2 - \sqrt3}{2 - \sqrt3}\\[1em] = \dfrac{2 - \sqrt3}{(2)^2 - (\sqrt3)^2 }\\[1em] = \dfrac{2 - \sqrt3}{4 - 3}\\[1em] = \dfrac{2 - \sqrt3}{1}\\[1em] = 2 - \sqrt3\\[1em] = 2 - 1.73\\[1em] = 0.27

Hence, 12+3\dfrac{1}{2 + \sqrt3} = 0.27.

(iii) Since, the denominator = 32−233\sqrt2 - 2\sqrt3, its rationalizing factor = 32+233\sqrt2 + 2\sqrt3.

432−23=432−23×32+2332+23=122+83(32)2−(23)2=122+8318−12=122+836=62+433=6×1.41+4×1.733=8.46+6.923=15.383=5.12\dfrac{4}{3\sqrt2 - 2\sqrt3} = \dfrac{4}{3\sqrt2 - 2\sqrt3} \times \dfrac{3\sqrt2 + 2\sqrt3}{3\sqrt2 + 2\sqrt3}\\[1em] = \dfrac{12\sqrt2 + 8\sqrt3}{(3\sqrt2)^2 - (2\sqrt3)^2}\\[1em] = \dfrac{12\sqrt2 + 8\sqrt3}{18 - 12}\\[1em] = \dfrac{12\sqrt2 + 8\sqrt3}{6}\\[1em] = \dfrac{6\sqrt2 + 4\sqrt3}{3}\\[1em] = \dfrac{6 \times 1.41 + 4 \times 1.73}{3}\\[1em] = \dfrac{8.46 + 6.92}{3}\\[1em] = \dfrac{15.38}{3}\\[1em] = 5.12

Hence, 432−23\dfrac{4}{3\sqrt2 - 2\sqrt3} = 5.12.

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