if $n$ is natural odd number then the polynom : $P(x)=x^n+ax^2+b$ has at the most 3 different roots

Question

I have this problem :

if $n$ is natural odd number then the polynom : $P(x)=x^n+ax^2+b$ has at the most 3 different roots.

$$P(x)=x^n+ax^2+b$$ $$P'(x)=nx^{n-1}+2ax$$ $$P''(x)=n(n-1)x^{n-2}+2a$$

I understand that I need to use Rolle's theorem in order to proof it, but I don't understand why P''(x) has only one root? why doesn't P''(x) has more roots than one?

Any ideas?

Thanks!

Answer 1

We have: $$P''(x)=n(n-1)x^{n-2}+2a$$

But $x^{n-2}$ is strictly increasing, so $P''$ is strictly increasing, so $P''$ can have at most one zero. It also clearly has at least one root.

Answer 2

The second derivative can only have one real root as $n-2$ is odd, and every real number has a unique real $n-2$th root.

And the problem amounts to finding a real number such that $x^{n-2}$ equals $\frac{-2a}{n(n-1)}$. So the solution is $$\sqrt[n-2]{\frac{-2a}{n(n-1)}}$$ or $$\text{sign}(a) \, \sqrt[n-2]{\frac{|2a|}{n(n-1)}}$$ depending on ones conventions.

Answer 3

Equal $P'(x)=nx^{n-1}+2ax$ to $0$ and find two point. (Suppose $a\not=0$)

$x_1=0$ and $x_2= \sqrt[n-2]{-2a}$ (Minima and maxima, or maxima and minima because $n-2$ is odd)

Suppose $x_1>x_2$

We now that $lim_{x \to +\infty}P(x)=+\infty$ and $ lim_{x \to -\infty}P(x)=-\infty$ So $x_1$ is a maxima and $x_2$ is a minima and $P(x_1)>P(x_2)$ (in $(x_1,x_2)$ $P(x)$ is strictly decreasing)

1. If $P(x_1)>0$ and $P(x_2)<0$ there are a root $z_0 \in (-\infty,x_1)$, a root $z_1 \in (x_1,x_2)$, a root $z_2 \in (x_2,+\infty)$
2. If $P(x_1)>0$ and $P(x_2)>0$ there is a root $z_0 \in (-\infty,x_1)$
3. If $P(x_1)<0$ and $P(x_2)<0$ there is a root $z_0 \in (x_2,+\infty)$

Analog for $x_1<x_2$

If $x_1=x_2 → a=0$ So we have that $x=0$ is a flex (the unique one) for $P(x)=x^n+b$ so there is only one root $x_0=\sqrt[n]{-b}$