Question

If $x=\frac{4\sqrt{6}}{\sqrt{2}+\sqrt{3}}$. find the value of $\frac{x+2\sqrt{2}}{x-2\sqrt{2}}+\frac{x+2\sqrt{3}}{x-2\sqrt{3}}$, using the properties of proportion.

Answer 1

Given, $x=\frac{4\sqrt{6}}{\sqrt{2}+\sqrt{3}}$
It may be written as,
$x=\frac{2\sqrt{2}\times 2\sqrt{3}}{\sqrt{2}+\sqrt{3}}$
or $\frac{x}{2\sqrt{2}}=\frac{2\sqrt{3}}{\sqrt{2}+\sqrt{3}}\dots \left(i\right)$
By componendo and dividendo, we get
$\frac{x+2\sqrt{2}}{x-2\sqrt{2}}=\frac{2\sqrt{3}+\sqrt{2}+\sqrt{3}}{2\sqrt{3}-\sqrt{2}-\sqrt{3}}=\frac{3\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\dots \left(ii\right)$
Equation (i) may also be written as
$\frac{x}{2\sqrt{3}}=\frac{2\sqrt{2}}{\sqrt{2}+\sqrt{3}}$
By componendo and dividendo, we get
$\frac{x+2\sqrt{3}}{x-2\sqrt{3}}=\frac{2\sqrt{2}+\sqrt{2}+\sqrt{3}}{2\sqrt{2}-\sqrt{2}-\sqrt{3}}=\frac{3\sqrt{2}+\sqrt{3}}{\sqrt{2}-\sqrt{3}}\dots \left(iii\right)$
Adding equation (ii) and (iii), we get
$\frac{x+2\sqrt{2}}{x-2\sqrt{2}}+\frac{x+2\sqrt{3}}{x-2\sqrt{3}}=\frac{3\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\frac{3\sqrt{2}+\sqrt{3}}{\sqrt{2}-\sqrt{3}}=\frac{3\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}-\left(\frac{3\sqrt{2}+\sqrt{3}}{\sqrt{3}-\sqrt{2}}\right)=\frac{3\sqrt{3}+\sqrt{2}-3\sqrt{2}-\sqrt{3}}{\sqrt{3}-\sqrt{2}}=\frac{2\sqrt{3}-2\sqrt{2}}{\sqrt{3}-\sqrt{2}}=\frac{2(\sqrt{3}-\sqrt{2})}{\sqrt{3}-\sqrt{2}}=2$