Let $p(x)$ be a real coefficient polynomial. Assume that there exits $a\in \mathbb{R}$ s.t. $p(a)\neq 0$ but $p'(a) = p''(a) = 0$. Show that $p(x)$ has at least one non-real root.

This is a problem from today's exam, and actually, I already have a solution. However, I can't understand the solution, so I tried to do it my self in the following direction:

Assume that $p(x)$ is a monic polynomial. We can just set $p(x) = (x-a)^{3}q(x) + B$ for some $B\neq 0$. If $p$ has only real roots, we have $p(x) = (x-\alpha_{1})\cdots (x-\alpha_{n})$ for some $\alpha_{1}\leq \cdots \leq \alpha_{n}$. Then we have $(x-a)^{3}q(x)+B = (x-\alpha_{1})\cdots(x-\alpha_{n})$, and I can' proceed from here.

Here's the original solution:

Solution: Observe that if $q(z)$ is a real-rooted polynomial with distinct roots, then by Rolle’s theorem $q'(z)$ is also real-rooted (since it has degree one less than the degree of $q$) and has the property that between every two roots of $q'$ there is a root of $q$. Since polynomials with distinct roots are dense in the set of real-rooted polynomials, this implies that if $q$ is any real-rooted polynomial and $q'(z)$ has a double root at $z$ then $q(z) = 0$.

For the given polynomial $p'(z)$ has a double root at $a$, but $p(a) \neq 0$, so $p$ cannot be real-rooted.

(Original image here.)

  • $\begingroup$ What was the solution? $\endgroup$ – 4-ier Aug 22 '18 at 6:43
  • $\begingroup$ @4-ier I just added. $\endgroup$ – Seewoo Lee Aug 22 '18 at 6:44
  • $\begingroup$ Thanks. What about it do you have trouble with? $\endgroup$ – 4-ier Aug 22 '18 at 6:45
  • $\begingroup$ @4-ier I can't understand the solution at all, especially the second sentence. $\endgroup$ – Seewoo Lee Aug 22 '18 at 6:48
  • $\begingroup$ Ok, thanks. Wanted to see where you were at. $\endgroup$ – 4-ier Aug 22 '18 at 6:49

Honestly, I find the "original solution" to be quite hand-waving the density argument.

For an alternative direct proof, consider that translation along the $\,x\,$ axis preserves the nature of the roots (real vs. complex) of a real polynomial, so it can be assumed WLOG that $\,a=0\,$, then:

$$p(x)=a_nx^n+a_{n-1}x^{n-1}+\ldots+a_3x^3+a_0 \quad\quad \style{font-family:inherit}{\text{with}} \;\;a_n \ne 0\;\;\style{font-family:inherit}{\text{and}}\;\; p(0)=a_0 \ne0$$

If the roots of $\,p(x)\,$ are $\,x_i \ne 0\,$ then the polynomial having as roots $\,y_i=1 / x_i\,$ is:


By Vieta's relations $\,\sum_i y_i = 0\,$ and $\,\sum_{i \lt j} y_iy_j = 0\,$, so $\,\sum_i y_i^2 = \left(\sum_i y_i\right)^2-2\sum_{i \lt j} y_iy_j=0\,$. If all $\,y_i\,$ were real, that would imply $\,y_i=0\,$, but $\,a_n \ne 0\,$ so none of the roots can be $\,0\,$. It follows that not all $\,y_i\,$ can be real, and therefore not all of $\,x_i\,$ are real.

| cite | improve this answer | |
  • 1
    $\begingroup$ I think this solution is much better than the original one to understand! Thank you very much. $\endgroup$ – Seewoo Lee Aug 23 '18 at 1:48

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.