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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 $\color\red{\forall a\in\mathbb{C}}$.

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    $\begingroup$ 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
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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).

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  • $\begingroup$ 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
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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?

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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:

$$12x^2+6ax+8$$

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.

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