Roots of $x^2+bx+c=0$ with $|b|\leq 1$ and $|c|\leq 1$ have absolute value at most the Golden Ratio. Coincidence?

I noticed a possibly interesting coincidence. Consider the set of quadratics $$x^2 + bx + c$$ with $$|b| \leq 1$$ and $$|c| \leq 1$$. For all $$x_0$$ such that $$x_0^2 + bx_0 + c = 0$$ we have $$|x_0| \leq \frac{1 + \sqrt{5}}{2}$$ where the term on the right is the golden ratio. This bound is sharp since $$x^2 - x - 1 = 0$$ has the golden ratio as a solution.

This inequality can be proved by using the quadratic equation and making the obvious estimates.

Is there anything deeper going on here?

My friend suggested that there might be a story about eigenvalues lurking here. I'm curious to hear Math SE's thoughts. I'm interested in different ways of looking at this problem, generalizations, or even references to other people making this observation.

• What exactly is the problem ? Do you want to prove the inequality ? Or am I missing something ? – The Demonix _ Hermit Dec 22 '19 at 8:19
• Not a problem per se. Just a mathematical observation I'm wondering if someone can comment on. If I have to formulate it as a problem, I would say "Is this a coincidence, or is there something interesting going on here?" – Charles Hudgins Dec 22 '19 at 8:38
• I am sorry , but the way it stands currently , it would be closed quite soon . – The Demonix _ Hermit Dec 22 '19 at 8:39
• alright. Should I just close it now? There isn't more to this than just "can someone tell me / point me to something interesting about this?" If this isn't the place for that, then I'll delete the post. – Charles Hudgins Dec 22 '19 at 8:43
• Voted to close, but, I suggest not voting to delete as some users find an answer below helpful. – Kemono Chen Dec 22 '19 at 11:21

What is a pair of roots of a quadratic polynomial from the given family? It is a pair of numbers $$\alpha,\beta$$ such that $$|\alpha+\beta|\leq1$$ and $$|\alpha\beta|\leq1$$. Plotting and shading these regions on the real plane, we get the following solution region.

The region containing the origin bounded by blue, green and two black lines is the solution region. Your observation reflects the fact that this region lies in the square formed by the lines $$x=\phi, x=-\phi, y = \phi \ \ \&\ \ y = -\phi$$, where $$\phi$$ is the golden ratio.

• Your figure is pleasing to the eye. May I ask which software you applied? Kind regards – Carl Christian Dec 22 '19 at 9:53
• @CarlChristian, Desmos. – Martund Dec 22 '19 at 10:02

Indeed for the equation $$az^2+bz+c=0$$ where $$a,b,c\in \mathbb{C}$$ such that $$|a|=|b|=|c|=r$$ (or up to scale setting $$r=1$$), the roots are bounded by

$$\frac{\sqrt{5}-1}{2} \le |z| \le \frac{\sqrt{5}+1}{2}$$

which appeared as the problem 4994 of the issue December 2007 and solved in March 2008 at the solutions-problem section of School Science and Mathematics Association.

The loci of the roots by varying either one of the $$a=e^{i\alpha}$$, $$b=e^{i\beta}$$ or $$c=e^{i\gamma}$$ are shown below, notice each locus is confined in the annulus $$\phi^{-1} \le r \le \phi$$.

See an article zeros of polynomials with cyclotomic coefficients for your further interest.