Question

If $x-\frac{1}{x}=4$ then find the value of $x^2+\frac{1}{x^2}$.

Answer 1

Given,

$x-\frac{1}{x}=4$

Now squaring both sides we get,

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

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

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

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