Question

If $(x+\frac{1}{x})=4$, find the values of
$x^2+\frac{1}{x^2}$

Answer 1

It is given that;

$\Rightarrow$ $x+\frac{1}{x}=4$

Squaring both sides,

$\Rightarrow$ $(x+\frac{1}{x}{)}^2=16$

$\Rightarrow$ $x^2+\frac{1}{x^2}+2.x.\frac{1}{x}=16$

$\Rightarrow$ $x^2+\frac{1}{x^2}=16-2$

$\Rightarrow$ $x^2+\frac{1}{x^2}=14$