Question

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

Answer 1

$x+\frac{1}{x}=3$

Squaring both sides,

${(x+\frac{1}{x})}^2=(3{)}^2\Rightarrow x^2+\frac{1}{x^2}+2=9\Rightarrow x^2+\frac{1}{x^2}=9-2=7\therefore x^2+\frac{1}{x^2}=7$