Question

If $x+\frac{1}{6}=6$ then find the value of $x^2+\frac{1}{x^2}\text{and}{x}^4+\frac{1}{x^4}$

Answer 1

$x+\frac{1}{x}=6$

Squaring both sides , we get :

${(x+\frac{1}{x})}^2=6^2\Rightarrow x^2+\frac{1}{x^2}+2=36\therefore x^2+\frac{1}{x^2}=34$

Again squaring both sides , we get ,

${(x^2+\frac{1}{x^2})}^2={\left(34\right)}^2\Rightarrow x^4+\frac{1}{x^4}+2=1156\therefore x^4+\frac{1}{x^4}=1154$