We have $f=x^4+ax^3+4x^2+1\in\mathbb{C}[x]$ with $x_1,x_2,x_3,x_4\in\mathbb{C}$.

We need to prove that $\color\red{\forall a\in\mathbb{C}},f$ doesn't have all real roots. How can I begin to solve this exercise.

Here is what I've tried:

$$\sum_{k=1}^4 x_k^2< 0\Rightarrow\:f\:doesn't\:have\:all\:real\:roots$$

  • Therefore $$\left(\sum_{k=1}^4 x_k\right)^2-2\left(\sum_{1\leq k<i\leq 4}x_k x_i\right)=a^2-8$$

$\Rightarrow a^2-8<0\Rightarrow a\in(-\sqrt{8},\sqrt{8})$

But what I proved was that $f$ doesn't have all real roots for $a\in(-\sqrt{8},\sqrt{8})$. I don't have ideea how can I prove that $f$ doesn't have all real roots $\color\red{\forall a\in\mathbb{C}}$.

  • 6
    $\begingroup$ Start with the easy part. If $a\in \mathbb{C}\setminus \mathbb{R}$, can $f$ have a real zero? $\endgroup$ – Daniel Fischer Jun 23 '15 at 14:27
  • $\begingroup$ @hHhh there is $x_i$ is something wrong? $\endgroup$ – Lucas Jun 23 '15 at 14:28
  • $\begingroup$ @DanielFischer what you want to mean by a real zero? $\endgroup$ – Lucas Jun 23 '15 at 14:30
  • $\begingroup$ A zero (root, but I prefer the term zero) that is an element of $\mathbb{R}$ (which means it is real, as opposed to e.g. purely imaginary). $\endgroup$ – Daniel Fischer Jun 23 '15 at 14:32
  • $\begingroup$ @Lucas Daniel Fischer was asking you to have a look at the roots of $f$ in the case $a$ not in $ \mathbb{R}$... $\endgroup$ – Martigan Jun 23 '15 at 14:33

Since $0$ is not a root, we can as well consider the “reverse” polynomial $$ g(x)=x^4+4x^2+ax+1 $$ whose roots are the reciprocals of the roots of $f$.

If $g$ has four distinct real roots, its derivative must vanish in three points; now $$ g'(x)=4x^3+8x+a $$ and the second derivative must vanish in two distinct points; but $$ g''(x)=12x^2+8 $$ has no real root.

This also settles the case of multiple roots, of course (with just a bit of work).

  • $\begingroup$ Clear-cut and convincing, unlike the other incomplete answers so far $\endgroup$ – Ewan Delanoy Jun 23 '15 at 16:07
  • $\begingroup$ egreg this is not similar with derivative for multiple root of order 4? $\endgroup$ – Lucas Jun 23 '15 at 16:52
  • $\begingroup$ @Lucas It's impossible for the equation to have a root of multiplicity $4$. $\endgroup$ – egreg Jun 23 '15 at 16:53
  • $\begingroup$ @egreg I don't understand how you figure out that "If g has four distinct real roots, its derivative must vanish in three points;" $\endgroup$ – Lucas Jun 23 '15 at 16:55
  • $\begingroup$ @Lucas Rolle's theorem applied to the interval between two consecutive roots. $\endgroup$ – egreg Jun 23 '15 at 16:59

Consider the case $a>0$. Then $f$ is increasing on $[0,\infty)$, so there are no positive zeros. Thus, if $f$ had four real zeros, they would have to be negative, and then by Rolle's theorem, $f'$ would have to have three negative zeros. But $f'(0)=0$, so...

Can you take it from there?


This is just a start.

If $a$ is complex, there are no real zeros. So we'll assume $a$ real.

Given any polynomial with $n$ real zeros, $f'(x)$ has at least $n-1$ real zeros, $f''(x)$ has $n-2$ real zeros, etc.

In this case, the second derivative is:


which, by dividing by by $3$, has the same number of roots as:

$$4x^2+2ax+\frac{8}{3}= \left(2x+\frac{a}{2}\right)^2 + \frac{8}{3}-\frac{a^2}{4}$$

So you need $\frac{a^2}{4}>\frac{8}{3}$ or $|a|>\sqrt{\frac{32}{3}}$ to have two real roots for the second derivative.


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.