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

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

From the given quadratic,
$x^2+1 = 13x$

Thus,
$x^2+\frac{1}{x^2} = \left(\frac{13x}{x}\right)^2 -2 = \boxed{167}$

x^2-13x+1=0 x^2+1=13x x^2+1/x=13 after simplification we get x+1/x=13 and on squaring both the sides we get x^2+1/x^2=167

Note that

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

Hence all we have to do is to find $x+\frac{1}{x}$ .

Since $x$ is not equal to $0$ , we can divide both sides of $x^{2}-13x+1=0$ by $x$ , we have

$x-13+\frac{1}{x}=0 \implies x+\frac{1}{x}=13$

Therefore, the desired answer is $13^{2}-2=\boxed{167}$ , and we are done.

on looking at eqn , product of roots is 1.. so we can assume it as x & 1/x, also x+1/x=13, put in formula of (x+1/x)^2=x^2+1/x^2 + 2, we get directly the answer...167