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I'm recalling this question from memory, so I may be messing it up a bit.

Let $a/3+b/2+c=0$. Show that $ax^2+bx+c=0$ has at least one root in $[0,1]$ using the Mean Value Theorem.

Let $f(x)=ax^2+bc+c$. Then $f(0)=c$ and $f(1)=a+b+c$. Also $f'(x)=2ax+b$. So there exists $f(\xi)=[f(1)-f(0)]/1=a+b-c$. Then $a+b-c=2a\xi+b \Rightarrow (a-c)/2a=\xi$.

I'm not sure if this is right or where to go from here.

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Just a quick note: you presumably mean $f(1) = a + b + c$. – JavaMan Dec 11 '12 at 6:15
@ortl: typo in third paragraph $f(x)=ax^2+bx+c$ (not $f(x)=ax^2+bc+c$) and $f(1)-f(0)=a+b$. – user103828 May 1 '14 at 6:27
up vote 3 down vote accepted

First, if $a =0$, then we have $bx + c = 0 \implies x = - \frac{c}{b} = \frac{b/2}{b} = \frac{1}{2}$.

Now, suppose $a \neq 0$. Note that $c = - \frac{a}{3} - \frac{b}{2}$, so you want to prove that the function $f(x) = ax^2 + bx - \frac{a}{3} - \frac{b}{2}$ has a root in $[0,1]$. We have $f(0) = - \frac{a}{3} - \frac{b}{2}$ and $f(1) = \frac{2}{3} a + \frac{1}{2} b$. Note that $$f(0)\cdot f(1) = - \frac{2}{9}a^2 - \frac{1}{4}b^2 < 0$$ as $a \neq 0$. Thus, $f(0)$ and $f(1)$ have different signs, and by the Intermediate Value Theorem, there is a root in $[0,1]$.

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Two quick comments when I look back at my answer. This is false if $a = b =0$ and $c \neq 0$. Also, my method was to use the Intermediate Value Theorem rather than the Mean Value Theorem, so to may not be quite the answer you were looking for. – JavaMan Dec 11 '12 at 16:04
$f(0)\cdot f(1) = - \frac{2}{9}a^2 - \frac{1}{2}ab - \frac{1}{4}b^2$, so $f(0)$ and $f(1)$ don't necessarily have different signs. – user103828 May 1 '14 at 6:46

Apply MVT to $g(x) = \int (ax^2+bx+c) dx$.

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I don't follow. My analysis class stopped at differentiation, so I can't utilize the FTC. – emka Dec 10 '12 at 21:55
@jdla: $\int(ax^2+bx+c)dx=\frac{a}{3}x^3+\frac{b}{2}x^2+cx$ – P.. Dec 10 '12 at 21:57
Perhaps you want to use a definite integral to define $g$. – David Mitra Dec 10 '12 at 22:06
Actually...would this be "cheating." In my analysis class I still don't have the foresight of knowing what integration is. – emka Dec 10 '12 at 22:07
Does this prove that there is a unique root in $[0,1]$? – Tom Oldfield Dec 10 '12 at 22:11

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