Question

If $x=3-2\sqrt{2}$,

find $x^4+\frac{1}{x^4}$

Answer 1

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

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

$=\frac{3+2\sqrt{2}}{1}$

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

$=3-2\sqrt{2}+3+2\sqrt{2}$

$=6$

Squaring on both sides,

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

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

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

Squaring on both sides,

${(x^2+\frac{1}{x^2})}^2=(34{)}^2$

$x^4+\frac{1}{x^4}=1156-2$

$\Rightarrow x^4+\frac{1}{x^4}=1154.$