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.



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 $-5<p<-4$.

  • $\begingroup$ Aye, thanks, I had a brainfart, reached the $-p-2$ step. Don't even how I could go wrong. $\endgroup$ – John Doe Apr 18 '15 at 15:39
  • 3
    $\begingroup$ The roots are real and distinct. That imposes the additional inequality $p^2-(-4p)\gt 0$. $\endgroup$ – André Nicolas Apr 18 '15 at 15:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.