Question

If $x=2\sqrt{3}+2\sqrt{2}$ and ${(x+\frac{1}{x})}^2$ is equal to $\frac{93+30\sqrt{6}}{m}$, then value of $m$ is?

Answer 1

Given, $x=2\sqrt{3}+2\sqrt{2}$.

Then, $\frac{1}{x}=\frac{1}{2\sqrt{3}+2\sqrt{2}}=\frac{2\sqrt{3}-2\sqrt{2}}{(2\sqrt{3}+2\sqrt{2})(2\sqrt{3}-2\sqrt{2})}$

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

$=\frac{2\sqrt{3}-2\sqrt{2}}{12-8}$ $=\frac{2(\sqrt{3}-\sqrt{2})}{4}$.

$\therefore \frac{1}{x}=\frac{(\sqrt{3}-\sqrt{2})}{2}$.

Then, $({x+\frac{1}{x})}^2=((2\sqrt{3}+2\sqrt{2})+\left(\frac{\sqrt{3}-\sqrt{2}}{2}\right){)}^2$

$=(\frac{5\sqrt{3}}{2}+\frac{3\sqrt{2}}{2}{)}^2=(\frac{5\sqrt{3}+3\sqrt{2}}{2}{)}^2$

$=\frac{75+30\sqrt{6}+18}{2}$ $=\frac{(93+30\sqrt{6})}{4}$.

On comparing, the value of $m$ is $4$.