TITLE: Show this equation has at least one root in $(0,1)$
QUESTION [7 upvotes]: Let $ax^2+bx+c=0$ be a quadratic equation, where $a,b,c\in\mathbb{R}$. If $2a+3b+6c=0$, then show that this equation will have atleast one root in $(0,1)$.
I think it involves either Rolle's Theorem or Lagrange's Mean Value Theorem, but can't think further. Please help, and yes, thanks in advance!

REPLY [3 votes]: By Viète from $2a+3b+6c=0$ we have $x_1x_2 - \frac{1}{2}(x_1+x_2)+\frac{1}{3}=0$, whence 
$(x_1 - \frac{1}{2})(x_2 - \frac{1}{2})=-\frac{1}{12}$.
But $|(x_1 - \frac{1}{2})(x_2 - \frac{1}{2})|\ge\frac{1}{4} $ if the roots are out (0,1).