Question

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

Answer 1

We have,

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

Since,

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

Therefore,

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

${(x+\frac{1}{x})}^2=4$

$x+\frac{1}{x}=\pm 2$

Hence, this is the answer.