The equation $x^2+px-p$ has two real and different solutions $x_1$ and $x_2$ for which $1<(x_1/x_2+x_2/x_1)<3$ for which interval of $p$? This is a multiple choice question, but I figure it will be  easy enough to do without the given answers.
It seems like an easy question. Use Vieta's formula from the inequality and define $p$ from that. But I'm making some stupid mistake somewhere that leads me to get an interval of $(-1,1)$ or something similar that isn't an option. So I'm looking for my mistake, but I can't find it.
Thanks!

By Vieta's formulas, one has
$$x_1+x_2=-\frac p1=-p,\ \ x_1x_2=\frac{-p}{1}=-p.$$
Since
$$\frac{x_1}{x_2}+\frac{x_2}{x_1}=\frac{(x_1+x_2)^2-2x_1x_2}{x_1x_2}=\frac{(-p)^2-2(-p)}{-p}=-p-2,$$
one has
$$1\lt -p-2\lt 3.$$
Added : Since $x_1\not=x_2$, one also needs $p^2-4(-p)\gt 0$. Thus, the answer will be $\color{red}{-5\lt p\lt -4}$.
