Question

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

Answer 1

Given : $x=2\sqrt{3}+2\sqrt{2}$

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

$=\frac{2\sqrt{3}-2\sqrt{2}}{12-8}$

$=\frac{2\sqrt{3}-2\sqrt{2}}{4}$

$x+\frac{1}{x}=2\sqrt{3}+2\sqrt{2}+\frac{2\sqrt{3}-2\sqrt{2}}{4}$

$=\frac{10\sqrt{3}+6\sqrt{2}}{4}$

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

$=\frac{5\sqrt{3}+3\sqrt{2}}{2}$.