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!

Since $x_1$ and $x_2$ are the roots of $x^2+px-p$ we can write, due to Vieta's formulas
\begin{align*}
\frac{x_1}{x_2}+\frac{x_2}{x_1}&=\frac{x_1^2+x_2^2}{x_1x_2}\\
&=\frac{(x_1+x_2)^2-2x_1x_2}{x_1x_2}\\
&=\frac{(-p)^2-2(-p)}{-p}\\
&=-p-2
\end{align*}
Then $1<x_1/x_2+x_2/x_1<3\quad\iff\quad -5<p<-3$. 

Also, being $x_1$ and $x_2$ real and distinct numbers we have $p^2-4(-p)>0$, hence $-5<p<-4$.
