Question

Find the value of $x^2+\frac{1}{x^2}$ if $x=5+2\sqrt{6}$

Answer 1

$x=5+2\times \sqrt{6}$

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

$=(5+2\times \sqrt{6}{)}^2+\frac{(5-2\times \sqrt{6}{)}^2}{(25-24{)}^2}$

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

$=(5+2\times \sqrt{6}{)}^2+(5-2\times \sqrt{6}{)}^2$

$=\left(25\right)+(2\times \sqrt{7}{)}^2$

$=49$