Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

if a is real, what is the only real number that could be a mutiple root of $x^3 +ax+1$=0

No one in my class know how to do it, so i have to ask it here.

share|cite|improve this question
Can we use calculus? – azarel Feb 19 '12 at 3:15
up vote 8 down vote accepted

Let the multiple root be $r$, and let the other root be $s$. If $r$ is to be real, then $s$ must be real also. From Vieta's formulas, we have $2r + s = 0$ and $r^2s = -1$. The first equation gives $s = -2r$, which we plug into the second equation to get $r^2s = -2r^3 = -1$, so $r = \boxed{\left(\frac12\right)^{1/3}}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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