Question

If $x=2+\sqrt{3}$, find the values of $x^2+\frac{1}{x^2}$.

Answer 1

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

$=2-\sqrt{3}$

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

$=4+3+4\sqrt{3}+4+3-4\sqrt{3}$

$=14$ Ans