TITLE: Show that $p'(c)+100p(c)=0$
QUESTION [7 upvotes]: Let $p(x)$ be a polynomial with real coefficients such that $a\le c \le b$ where $a,b$ are two consecutive roots of $p(x)$. Show that there exists at least one c for which $$p'(c)+100p(c)=0$$

Okay, so this is what I tried, By Lagrange's Mean Value Theorem we can find a point $c$ in the interval $[a,b]$ such that $p'(c)=0$ but at that time, $100p(c)$ is not $0$. I tried some other ways but I don't think those will work. Any help is appreciated.

REPLY [4 votes]: $p(x)$ is a smooth function, thus $p'(x) + 100p(x)$ is a smooth function as well. 
Case 1: $0 \ne \text{sign}(p'(a)) \neq \text{sign}(p'(b)) \neq 0$. Thus two points exist where the function is positive and negative: $p'(a) + 100p(a) < 0$ and  $p'(b) + 100p(b) > 0$ (or vice versa). As the function is smooth, it is continuous everywhere, hence it has to assume $0$ between $a$ and $b$.
Case 2: $\text{sign}(p'(a)) = 0$ or  $\text{sign}(p'(b)) = 0$. In this case $p'(a) + 100p(a) = 0$ or $p'(b) + 100p(b) = 0$ as $a$ and $b$ are roots.