Quadratic root in $(0,1)$

 A nice little problem from the IIT JEE exam.

$a,b,c$ are real numbers such that $$2a + 3b + 6c = 0$$

Show that $ax^2 + bx + c = 0$ has a root in the interval $(0,1)$

Scroll down for a short solution.

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Solution:

Consider $$ f(x) = 2a x^3 + 3bx^2 + 6cx + d$$

We have that $f(0) = d$ and $f(1) = d$. Thus by Rolle's theorem, there is some $\eta \in (0,1)$ such that $f'(\eta) = 0$.

Now $f'(x) = 6ax^2 + 6bx + 6c = 6(ax^2 + bx + c)$. 

Thus $a\eta^2 + b\eta + c = 0$ and we are done!