Question

If $x=3+\sqrt{8}$, then find the value of $x^2+\frac{1}{x^2}$

Answer 1

We have,

$x=3+\sqrt{8}$

Here,

$x^2+\frac{1}{x^2}$

$={(3+\sqrt{8})}^2+\frac{1}{{(3+\sqrt{8})}^2}$

$=9+8+6\sqrt{8}+\frac{1}{9+8+6\sqrt{8}}$

$=17+6\sqrt{8}+\frac{1}{17+6\sqrt{8}}$

$=17+6\sqrt{8}+\frac{1}{(17+6\sqrt{8})}\times \left(\frac{17-6\sqrt{8}}{17-6\sqrt{8}}\right)$

$=17+6\sqrt{8}+\frac{17-6\sqrt{8}}{({17}^2-{\left(6\sqrt{8}\right)}^2)}$

$=17+6\sqrt{8}+\frac{17-6\sqrt{8}}{289-288}$

$=17+6\sqrt{8}+\frac{17-6\sqrt{8}}{1}$

$=17+6\sqrt{8}+17-6\sqrt{8}$

$=34$

Hence, the value is $34$.