Prove that between two roots of $f(x)$ there is a root of $g(x)$ Let $f(x),g(x)$ be differential functions, and $f'(x)g(x)\neq f(x)g'(x)$ for all $x\in\mathbb R$. Prove that between two roots of $f(x)$ there is a root of $g(x)$.

I guess this has to do with Rolle's theorem. I saw that when $f'(x)=0$, $g(x)\neq0$ and when $f(x)=0$, $g'(x)\neq0$, but I didn't manage to prove the conjecture. Thanks for any help!
Suppose that $a,b$ are roots of $f$ with $a&lt;b$ and $g(x)\neq0$ for $x\in (a,b)$. Consider the function $h(x)=\dfrac{f(x)}{g(x)}$ in $[a,b]$ to derive a contradiction(is well defined, differentiable on [a,b] with $h(a)=h(b)$...).
If $g(x) \neq 0$ for all $x \in (a,b)$ then $h(x)=\frac{f(x)}{g(x)}$ is well definied and differentiable. Also $h’(x)=\frac{fg’-f’g}{g^2}=0$ by hypothesis and $h(x)=c$ with $c \in \mathbb{R}\setminus\{0\}$ and $f(a)=cg(a)=0=cg(b)$ then $g(a)=0= g(b)$ contradiction 
Sorry, my english is too bad. 
Must be $fg’-f’g=0$ and $a,b$ consecutive roots of $f$ and $g(a)$ and $g(b)$ not both zero