# Existence of real roots of a quartic polynomial

Question

What is the minimum possible value of $a^{2}+b^{2}$ so that the polynomial $x^{4}+ax^{3}+bx^{2}+ax+1=0$ has at least 1 root?

Attempt

I divided by $x^{2}$ and got $$x^{2}+\frac{1}{x^{2}}+2+a\left(x+\frac{1}{x}\right)+b-2=0$$ by letting $$x+\frac{1}{x}=X$$ the equation becomes: $$X^{2}+aX+(b-2)=0$$ $$\therefore X=\frac{-a\pm \sqrt{a^{2}-4b+8}}{2}$$ but I am not sure how to continue. If the polynomial has 1 root doesn't that is should be a double root? Are we counting multiplicity or not?

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I think you may be thinking that if $x$ is a root, then so is $\frac 1 x$, so roots indeed come in pairs (with some special cases to consider). Once you have found $X$ you still have a quadratic to solve for $x$, so you need a condition on $X$ which makes $x$ real. – Mark Bennet Apr 27 '13 at 18:07
"At least one root" means it should have one, not excluding it has more (or one of them is double). – egreg Apr 27 '13 at 19:43
for any future readers, another solution can be found on this link. I unfortunately posted the same question two years later math.stackexchange.com/questions/1439501/… – Rishi Sep 18 '15 at 1:50

It seems to me that requesting the smallest possible value for $a^2+b^2$ is almost a red herring.

Claim.

For $|a| \leq 4$ and $b > 2 |a| - 2$, the polynomial $$p(x) = x^4 + ax^3 + bx^2 + ax + 1$$ has no real zeros.

Proof.

First set $b = 2|a| - 2 + \epsilon$, where $\epsilon > 0$. If we denote $\pm = \operatorname{sign} a$, so that $a = \pm |a|$, we have

\begin{align} p(x) &= x^4 + ax^3 + (2|a|-2+\epsilon)x^2 + ax + 1 \\ &= x^4 - 2x^2 + 1 + x(ax^2 + 2|a|x + a) + \epsilon x^2 \\ &= (x^2-1)^2 + ax(x \pm 1)^2 + \epsilon x^2 \\ &= (x \pm 1)^2\left[ (x \mp 1)^2 + ax \right] + \epsilon x^2 \\ &= (x \pm 1)^2\left[ x^2 \pm (|a|-2)x + 1 \right] + \epsilon x^2 \\ &\geq (x \pm 1)^2 \min\left\{x^2 - 2x + 1,x^2+2x+1\right\} + \epsilon x^2 \\ &= (x \pm 1)^2\min\left\{(x-1)^2,(x+1)^2\right\} + \epsilon x^2 \\ &> 0. \end{align}

Q.E.D.

The points on the boundary of the region

$$A = \{(a,b) \,\colon |a| \leq 4 \,\,\,\text{and}\,\,\,b > 2 |a| - 2\}$$

which have the smallest norm are

$$(a,b) = \left(\pm \frac{4}{5},-\frac{2}{5}\right),$$

as shown in the following image.

It only remains to demonstrate that one of these points yields a polynomial with a real zero. In fact they both do:

$$x^4 + \frac{4}{5} x^3 - \frac{2}{5} x^2 + \frac{4}{5} x + 1$$

has a zero at $x=-1$ and

$$x^4 - \frac{4}{5} x^3 - \frac{2}{5} x^2 - \frac{4}{5} x + 1$$

has a zero at $x=1$.

Thus

The smallest value of $a^2 + b^2$ for which we can find some $a,b\in \mathbb R$ such that $p(x)$ has a real zero is $$\left(\frac{4}{5}\right)^2 + \left(\frac{2}{5}\right)^2 = \frac{4}{5}.$$

Below is a plot of the $(a,b)$-plane which shows the region (in blue) where the polynomial $p(x)$ has at least one real root. Note that for $|a| \leq 4$ the boundary of this region is precisely $b = 2|a| - 2$ (shown in black), but for $|a| > 4$ it curves inward.

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Thank you so much!!! – please delete me Apr 29 '13 at 15:36
@AlexanderJones you're very welcome. It was a fun problem. – Antonio Vargas Apr 29 '13 at 15:39
By the way, I found this problem in a mathematical olympiad paper, I tried to do it for so long but couldn't, I don't know how maths olympiad people can do this so quickly :( – please delete me Apr 29 '13 at 15:55
@AlexanderJones, if it makes you feel any better I wouldn't have been able to solve it in an olympiad setting either. The first thing I did was use my computer to plot the second image in my post. Only once I could roughly see what was going on was I able to construct my answer. – Antonio Vargas Apr 29 '13 at 16:00