Question

If $x+\frac{1}{x}=3$, then the value of $x^2+\frac{1}{x^2}=$

Answer 1

Given $(x+\frac{1}{x})=3$

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

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

$\Rightarrow x^2+\frac{1}{x^2}=3^2-2=7$