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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!

Ashish Gaurav
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    We can simplify by dividing by $a$ (in case $a\ne 0$), i.e. assuming $a=1$: $x^2+bx+c=0$ with $3b+6c=-2$. Then the case $a=0$ has to be considered separately, it yields $x=-c/b$, I bet it is in $(0,1)$ whenever $3b+6c=0$. – Berci Feb 19 '13 at 14:34
  • I didn't mention, but I think it is assumed that the quadratic equation never converts to linear (a is never 0) – Ashish Gaurav Feb 19 '13 at 14:40

3 Answers3

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Hint: Another way, beside's to @Berci's comment is to consider the function $$f(x)=\frac{1}{3}ax^3+\frac{1}{2}bx^2+cx$$ on $I=(0,1)$.

Brian M. Scott
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Mikasa
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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).

Boris Novikov
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I think I finally got it.
Using(as Babac suggested) $$f(x)=\frac{1}{3}ax^3+\frac{1}{2}bx^2+cx$$ Since this is continuous in $[0,1]$ and differentiable in $(0,1)$, so by Lagrange's Theorem
there exists at least one $c\in(0,1)$ such that $$f'(c)=\frac{f(1)-f(0)}{1-0}$$ This gives the result we need.

Ashish Gaurav
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