Question

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

Answer 1

$x=3+\sqrt{8}$

$\therefore \frac{1}{x}=\frac{1}{3+\sqrt{8}}=\frac{1(3-\sqrt{8})}{(3+\sqrt{8})(3-\sqrt{8})}$

(Rationalising the denominator)

$=\frac{3-\sqrt{8}}{(3{)}^2-(\sqrt{8}{)}^2}=\frac{3-\sqrt{8}}{9-8}=\frac{3-\sqrt{8}}{1}$

$=3-\sqrt{8}$

Now, $x+\frac{1}{x}=3+\sqrt{8}+3-\sqrt{8}=6$

Squaring both sides

${(x+\frac{1}{x})}^2=(6{)}^2$

$\Rightarrow x^2+\frac{1}{x^2}+2=36$

$\Rightarrow x^2+\frac{1}{x^2}=36-2=34$