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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 $f(1) = a + b + c$.2012-12-11
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    @ortl: typo in third paragraph $f(x)=ax^2+bx+c$ (not $f(x)=ax^2+bc+c$) and $f(1)-f(0)=a+b$.2014-05-01

2 Answers 2

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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.2012-12-11
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    $f(0)\cdot f(1) = - \frac{2}{9}a^2 - \frac{1}{2}ab - \frac{1}{4}b^2$, so $f(0)$ and $f(1)$ don't necessarily have different signs.2014-05-01
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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.2012-12-10
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    @jdla: $\int(ax^2+bx+c)dx=\frac{a}{3}x^3+\frac{b}{2}x^2+cx$2012-12-10
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    @jdla You don't need the fundamental theorem of calculus, as you can calculate the integral *explicitly*, as Pambos points out.2012-12-10
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    Ah sorry. I was for some reason viewing it as the $\int f'(x)$.2012-12-10
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    Perhaps you want to use a definite integral to define $g$.2012-12-10
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    Actually...would this be "cheating." In my analysis class I still don't have the foresight of knowing what integration is.2012-12-10
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    Does this prove that there is a unique root in $[0,1]$?2012-12-10
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    @TomOldfield Surely not, but the "exactly one root" bit is wrong in the first place. Counterexample: $6x^2-6x+1=0$, whose two roots are $\frac12\pm\frac{1}{\sqrt{12}}\in[0,1]$.2012-12-10
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    @jdla About cheating, ... this is a good question that I have no answer, but +1 for your academic integrity.2012-12-10
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    @user1551: The OP changed his post to "at least one real root" so it looks like your answer should do the trick.2012-12-11