Prove that there is $c \in (a,b)$ for which $g(c) = 0$ 
Suppose $f$ and $g$ are differentiable functions over an open interval $I$ satisfying $f'g = fg'$. Assume $a<b$ are adjacent roots of $f(x) = 0$ for some $a,b \in I$ and $g(a)g(b) \neq 0$. Prove that there is $c \in (a,b)$ for which $g(c) = 0$. 

Attempt
We have that $f(x) = (x-a)(x-b)Q(x)$ and also that $$[(x-a)(x-b)Q'(x)+Q(x)((2x-(a+b))]g(x) = (x-a)(x-b)Q(x)g'(x).$$ I am not sure if this helps or not but I think Rolle's theorem might definitely help for this question.
 A: $a,b$ are adjacent roots, implies $f\neq 0$ on $(a,b)$, suppose that $g\neq 0$ on $(a,b)$, on $(a,b)$ we have $f'/f=g'/g$ thus $log\mid f\mid =log\mid g\mid +c$ this implies that $f=Cg$ where $C$ is a non zero constant on $(a,b)$, this implies that $f=Cg$ on $[a,b]$, thus $g(a)=g(b)=0$. Contradiction.
A: Let us assume that $g(c) \neq 0$ for all $c \in (a, b)$. Given that $g(a)g(b) \neq 0$ it follows that $g(x) \neq 0$ for all $x \in [a, b]$. And hence the function $F(x) = f(x)/g(x)$ is defined for all $x \in [a, b]$. Clearly we have $$F'(x) = \frac{g(x)f'(x) - f(x)g'(x)}{\{g(x)\}^{2}} = 0$$ and hence $F(x) = k$ for all $x \in [a, b]$ where $k$ is some constant. Therefore $f(x) = kg(x)$ and clearly $k \neq 0$ otherwise $f$ will be identically $0$ (note that the problem says that $a, b$ are adjacent roots of $f(x) = 0$ and hence it means that $f(x) \neq 0$ for all $x \in (a, b)$).
It thus follows that $g(x) = f(x)/k$ and clearly $g(a)g(b) = f(a)f(b)/k^{2} = 0$ which is a contradiction. Hence our assumption that $g(c) \neq 0$ for all $c \in (a, b)$ is wrong. And therefore there is at least one $c \in (a, b)$ for which $g(c) = 0$.
