Question

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

Answer 1

${\displaystyle a=\frac{4\sqrt{6}}{\sqrt{2}+\sqrt{3}}}$

${\displaystyle \frac{a}{2\sqrt{2}}=\frac{2\sqrt{3}}{\sqrt{2}+\sqrt{3}}}$

Applying componendo and dividendo,

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

${\displaystyle \frac{a+2\sqrt{2}}{a-2\sqrt{2}}=\frac{3\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}}$  ...(1)

${\displaystyle \frac{a}{2\sqrt{3}}=\frac{2\sqrt{2}}{\sqrt{2}+\sqrt{3}}}$

Applying componendo and dividendo,

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

${\displaystyle \frac{a+2\sqrt{3}}{a-2\sqrt{3}}=\frac{3\sqrt{2}+\sqrt{3}}{\sqrt{2}-\sqrt{3}}}$  ...(2)

From (1) and (2),

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

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

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