Question

If $x=5-2\sqrt{6},$ find $x^2+\frac{1}{x^2}$.

Answer 1

We know, $x^2+\frac{1}{x^2}={(x-\frac{1}{x})}^2+2$

Given, $x=5-2\sqrt{6}$.

Then, $\frac{1}{x}=\frac{1}{5-2\sqrt{6}}=\frac{5+2\sqrt{6}}{(5-2\sqrt{6})(5+2\sqrt{6})}=\frac{5+2\sqrt{6}}{5^2-(2\sqrt{6}{)}^2}=\frac{5+2\sqrt{6}}{25-24}=\frac{5+2\sqrt{6}}{1}=5+2\sqrt{6}$.

Then, $x^2+\frac{1}{x}=(5-2\sqrt{6}-(5+2\sqrt{6}){)}^2+2$

$\therefore {(-4\sqrt{6})}^2+2=96+2=98$.

Therefore, the required value is $98$.