Question

If $x=2\sqrt{3}+2\sqrt{2}$, then find the value of : ${(x+\frac{1}{x})}^2.$

Answer 1

Given that,

$x=2\sqrt{3}+2\sqrt{2}=2(\sqrt{3}+\sqrt{2})$

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

$=\frac{\sqrt{3}-\sqrt{2}}{2\left[\right(\sqrt{3}{)}^2-\left(\sqrt{2}{)}^2\right]}=\frac{\sqrt{3}-\sqrt{2}}{2(3-2)}=\frac{\sqrt{3}-\sqrt{2}}{2}$

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

$=\frac{{(5\sqrt{3}+3\sqrt{2})}^2}{4}$

$=\frac{75+18+30\sqrt{6}}{4}$

$=\frac{93+30\sqrt{6}}{4}$