6
$\begingroup$

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.

  • 0
    @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

3

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 $[0,1]$.

  • 0
    $f(0)\cdot f(1) = - \frac{2}{9}a^2 - \frac{1}{2}ab - \frac{1}{4}b^2$, so $f(0)$ and $f(1)$ don't necessarily have different signs.2014-05-01
5

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

  • 0
    @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