Question

If $x^2+\frac{1}{x^2}=\frac{17}{4}$ then find the value of $x+\frac{1}{x}$

Answer 1

$x^2+\frac{1}{x^2}=\frac{17}{4}\Rightarrow x^2+\frac{1}{x^2}+2-2=\frac{17}{4}\Rightarrow {(x+\frac{1}{x})}^2=\frac{17}{4}+2\Rightarrow {(x+\frac{1}{x})}^2=\frac{25}{4}$
$\therefore x+\frac{1}{x}=\frac{5}{2},\frac{-5}{2}$