Question

If $a=\frac{4\sqrt{6}}{\sqrt{2}+\sqrt{3}}$, find the value of
$\frac{a+2\sqrt{2}}{a-2\sqrt{2}}+\frac{a+2\sqrt{3}}{a-2\sqrt{3}}$

Answer 1

$a=\frac{4\sqrt{6}}{\sqrt{2}+\sqrt{3}}=\frac{4\sqrt{6}}{\sqrt{2}+\sqrt{3}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}=4\sqrt{6}(\sqrt{3}-\sqrt{2})$

$⟹a=4\sqrt{18}-4\sqrt{12}=12\sqrt{2}-8\sqrt{3}$

$\frac{a+2\sqrt{2}}{a-2\sqrt{2}}+\frac{a+2\sqrt{3}}{a-2\sqrt{3}}$

$=\frac{14\sqrt{2}-8\sqrt{3}}{10\sqrt{2}-8\sqrt{3}}+\frac{12\sqrt{2}-6\sqrt{3}}{12\sqrt{2}-10\sqrt{3}}$

$=1+\frac{2\sqrt{2}}{5\sqrt{2}-4\sqrt{3}}+1+\frac{2\sqrt{3}}{6\sqrt{2}-5\sqrt{3}}$

On Rationalising we get-

$=1+\sqrt{2}(5\sqrt{2}+4\sqrt{3})+1-\frac{2\sqrt{3}}{3}(6\sqrt{2}+5\sqrt{3})$

$=1+10+4\sqrt{6}+1-4\sqrt{6}-10$

$=2$