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Let $$P(x)=x^4+px^3+qx^2+px+1$$ where $p,q \in \mathbb{R}$.

Let $\mathcal{R}$ be the region in the $pq$ - plane such that $P(x)$ has no real roots. Find the region $\mathcal{R}$.

My Progress:

Using the substitution $u=x+\frac{1}{x}$, we can reduce $P(x)$ to become $Q(u)=u^2+pu+(q-2)$. Since $P(x)$ has no real roots, then $Q(u)$ has no real roots ie: $\Delta < 0$. From this, we get

$$q > \frac{p^2}{4}+2$$

Problem:

Clearly $(p,q)=(0,0)$ will result in $P(x)=0$ having no real roots. But this point is not covered in the region I have found above. So, I am suspecting that there is another bounding region that I am missing.

Questions:

  1. What is this missing region?
  2. Can I somehow use the fact that if $x \in \mathbb{R}$, then $\left | x+\frac{1}{x} \right | \geq 2$ ?
  3. $P(x)$ is a symmetric polynomial, so if $\alpha$ is a root, then $\frac{1}{\alpha}$ is also a root. Can this be used?
Community
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Trogdor
  • 10,331

2 Answers2

1

But $Q(u)$ can have real roots. If both the roots of $Q(u)=0$ lie in the interval $(-2,2)$ then $x+1/x=u$ is insoluble over $\Bbb R$ and $P(x)=0$ will have no real roots. So you need to find all $p$, $q$ such that either $Q(u)=0$ has no real roots, or two real roots both in $(-2,2)$.

Angina Seng
  • 158,341
0

you have to consider the equation $$x+\frac{1}{x}=-\frac{p}{2}\pm\sqrt{\frac{p^2}{4}+2-q}$$ and solve this for $x$ and you will get an additional condition.

Dr. Sonnhard Graubner
  • 95,283