Question

Simplified form after rationalising the denominator: $\frac{7\sqrt{3}}{\sqrt{10}+\sqrt{3}}-\frac{2\sqrt{5}}{\sqrt{6}+\sqrt{5}}$ is $-\sqrt{m}+7$. Find $m$.

Answer 1

$\frac{7\sqrt{3}}{\sqrt{10}+\sqrt{3}}-\frac{2\sqrt{5}}{\sqrt{6}+\sqrt{5}}=\frac{7\sqrt{3}(\sqrt{6}+\sqrt{5})-2\sqrt{5}(\sqrt{10}+\sqrt{3})}{(\sqrt{10}+\sqrt{3})(\sqrt{6}+\sqrt{5})}$

$=\frac{21\sqrt{2}+7\sqrt{3}\sqrt{5}-10\sqrt{2}-2\sqrt{3}\sqrt{5}}{\sqrt{10}\sqrt{6}+\sqrt{10}\sqrt{5}+3\sqrt{2}+\sqrt{3}\sqrt{5}}$

$=\frac{11\sqrt{2}+5\sqrt{15}}{3\sqrt{15}+8\sqrt{2}}\times \frac{3\sqrt{15}-8\sqrt{2}}{3\sqrt{15}+8\sqrt{2}}$ (Rationalizing)

$=\frac{11\sqrt{2}(3\sqrt{15}-8\sqrt{2})+5\sqrt{15}(3\sqrt{15}-8\sqrt{2})}{9\times 15-64\times 2}$

$=\frac{33\sqrt{30}-88\times 2+15\times 15-40\sqrt{30}}{135-128}$

$=\frac{-7\sqrt{30}+225-176}{7}$

$=-\sqrt{30}+7=-\sqrt{m}+7$ (given)

$\therefore m=30$