Question

Simplify :

(i) $\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}+\frac{\sqrt{12}}{\sqrt{3}-\sqrt{2}}$

(ii) $\frac{7+3\sqrt{5}}{3+\sqrt{5}}-\frac{7-3\sqrt{5}}{3-\sqrt{5}}$

Answer 1

(i) $\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}+\frac{\sqrt{12}}{\sqrt{3}-\sqrt{2}}$

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

(Rationalising the denominator)

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

$=\frac{9\times 2+4\times 3-2\times 3\sqrt{2}\times 2\sqrt{3}}{9\times 2-4\times 3}$

$=\frac{18+12-12\sqrt{6}}{18-12}=\frac{30-12\sqrt{6}}{6}=5-2\sqrt{6}$

And, $\frac{\sqrt{12}}{\sqrt{3}-\sqrt{2}}=\frac{\left(\sqrt{4\times 3}\right)(\sqrt{3}+\sqrt{2})}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}$

$=\frac{2\sqrt{3}(\sqrt{3}+\sqrt{2})}{(\sqrt{3}{)}^2-(\sqrt{2}{)}^2}=\frac{2\times 3+2\sqrt{6}}{3-2}$

$=\frac{6+2\sqrt{6}}{1}=6+2\sqrt{6}$

$\therefore \frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}+\frac{\sqrt{12}}{\sqrt{3}-\sqrt{2}}=5-2\sqrt{6}+6+2\sqrt{6}=11$

(ii) $\frac{7+3\sqrt{5}}{3+\sqrt{5}}-\frac{7-3\sqrt{5}}{3-\sqrt{5}}$

$=\frac{(7+3\sqrt{5})(3-\sqrt{5})-(7-3\sqrt{5})(3+\sqrt{5})}{(3+\sqrt{5})(3-\sqrt{5})}$

$=\frac{(21-7\sqrt{5}+9\sqrt{5}-3\times 5)-(21+7\sqrt{5}-9\sqrt{5}-3\sqrt{5}\times \sqrt{5})}{(3{)}^2-(\sqrt{5}{)}^2}$

$=\frac{(21+2\sqrt{5}-15)-(21-2\sqrt{5}-15)}{9-5}$

$=\frac{(6+2\sqrt{5})-(6-2\sqrt{5})}{4}$

$=\frac{6+2\sqrt{5}-6+2\sqrt{5}}{4}=\frac{4\sqrt{5}}{4}=\sqrt{5}$