Question

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

Answer 1

$x=3+\sqrt{8}{x}^2={(3+\sqrt{8})}^2=9+8+(2\times 3\times \sqrt{8})=17+6\sqrt{8}\frac{1}{x}=\frac{1}{3+\sqrt{8}}=\frac{1}{3+\sqrt{8}}\frac{3-\sqrt{8}}{3-\sqrt{8}}=\frac{3-\sqrt{8}}{9-8}=3-\sqrt{8}\Rightarrow \frac{1}{x^2}={(3-\sqrt{8})}^2=9+8-2\times 3\times \sqrt{8}=17-6\sqrt{8}{x}^2+\frac{1}{x^2}=17+6\sqrt{8}+17-6\sqrt{8}=34$