Question

If $x^2-5x-1=0$, find the value of $x^2+\frac{1}{x^2}$

Answer 1

given $x^2-5x-1=0$

let a,b be roots

then $a+b=5,ab=-1$

so x can be $a,\frac{-1}{a}$

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

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

or

$\left(\right(\frac{-1}{a})-\frac{1}{\frac{-1}{a}}{)}^2+2$

$=5^2+2=27$