Question

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

Answer 1

Given that:

$x=2+\sqrt{3}$

then,

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

$x^2=4+4\sqrt{3}+3$

$x^2=7+4\sqrt{3}$

Then,

$x^2+\frac{1}{x^2}=(7+4\sqrt{3})+\frac{1}{7+4\sqrt{3}}$

$=(7+4\sqrt{3})+\frac{7-4\sqrt{3}}{(7+4\sqrt{3})(7-4sqrt3)}$

$=(7+4\sqrt{3})+\frac{7-4\sqrt{3}}{7^2-(4\sqrt{3}{)}^2}$

$=(7+4\sqrt{3})+\frac{7-4\sqrt{3}}{49-48}$

$=7+4\sqrt{3}+7-4\sqrt{3}$

$=14$

So,

$x^2+\frac{1}{x^2}=14$ Answer