Matematické Fórum

stuart clark · 26. 05. 2011 05:00

Maximum number of real solution of the equation $ax^n+x^2+bx+c=0$ Where $a,b,c\in\mathbb{R}$ and $n=$ even positive Integer.

musixx · 26. 05. 2011 10:03 — Editoval musixx (26. 05. 2011 10:07)

Let's investigate local extrems of a function $f(x)=ax^n+x^2+bx+c$ , and let's assume a nontrivial case $n>2$ and even.

Its first derivative is $f^\prime(x)=anx^{n-1}+2x+b$ .

A degree of a polynomial $anx^{n-1}$ is odd. Therefore there can be only one intersection with a line $-2x-b$ for positive $a$ , and up to three intersections with a negative $a$ . A case of positive $a$ is therefore not interesting, since it can lead only up to two real roots of $f(x)$ .

If we further analyze the second derivative $f^{\prime\prime}(x)=an(n-1)x^{n-2}+2$ ( $a$ negative), we can see that we can have a sequence (from left to right on axis x) local maximum -- local minimum -- local maximum which could mean four roots (one in each of four invervals determined by zeroes of $f^\prime(x)$ ).

In the same time this situation is achievable which is proven via a polynomial $-x^4+x^2-0.1$ (clearly a maximum of $-x^4+x^2$ is 1/4, and it touches axis x in zero, therefore the subtraction of 0.1 which is somewhere between 0 and 1/4).

Final answer: such a polynomial can contain up to four real roots.