Question

For $x=2\sqrt{3}+2\sqrt{2},$ if ${(x+\frac{1}{x})}^2$ is $\frac{93+30\sqrt{m}}{4}$ then $m$ is :

Answer 1

$x=2\sqrt{3}+2\sqrt{2}\frac{1}{x}=\frac{1}{2\sqrt{3}+2\sqrt{2}}$

We rationalize the denominator, we get

$\frac{1}{x}=\frac{1}{2\sqrt{3}+2\sqrt{2}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}=\frac{\sqrt{3}-\sqrt{2}}{2(3-2)}=\frac{\sqrt{3}-\sqrt{2}}{2}x+\frac{1}{x}=2\sqrt{3}+2\sqrt{2}+\frac{\sqrt{3}-\sqrt{2}}{2}=\frac{5\sqrt{3}+3\sqrt{2}}{2}$

${(x+\frac{1}{x})}^2=\left(\frac{5\sqrt{3}+3\sqrt{2}}{2}\right)\left(\frac{5\sqrt{3}+3\sqrt{2}}{2}\right)$

$=\frac{25\times 3+9\times 2+30\sqrt{6}}{4}=\frac{93+30\sqrt{6}}{4}$

Hence, $m=6$.