Question

If $x^2+\frac{1}{x^2}=23,$ evaluate $x+\frac{1}{x}$

Answer 1

$x^2+\frac{1}{x^2}=23$

To find: $x+\frac{1}{x}$

$\therefore {(x+\frac{1}{x})}^2=x^2+\frac{1}{x^2}+2[\therefore x^2+\frac{1}{x^2}=23]$

$\therefore {(x+\frac{1}{x})}^2=23+2$

$(x+\frac{1}{x}{)}^2=25$

$(x+\frac{1}{x})=\sqrt{25}$

$\therefore x+\frac{1}{x}=5$