# 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 $\color{red}{-5\lt p\lt -4}$.

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 $-5<p<-4$.

• Aye, thanks, I had a brainfart, reached the $-p-2$ step. Don't even how I could go wrong. – John Doe Apr 18 '15 at 15:39
• The roots are real and distinct. That imposes the additional inequality $p^2-(-4p)\gt 0$. – André Nicolas Apr 18 '15 at 15:54