Does a quadratic necessarily have a root in this interval? If F(x) is the quadratic $ax^2+bx+c$ with $ac>0$ $b^2-4ac>0$, it is true that within the interval $[-\frac{b}{a},+\frac{b}{a}]$ there exists a point $x$ where $F(x)=0$. I was told this earlier but I don't see how that is necessarily true.
 A: You can do what J.M. suggested, i.e. use the quadratic formula.
First, it's clear both a,c have the same sign, so multiplying by $\,\,-\frac{1}{a}\,\,$ if necessary we can assume the equation is $\,\,x^2+bx+c=0\,\,,\,\,c>0$ .
Using the quad. form we get that we must prove  $$\,\,\displaystyle{-\frac{|b|}{2}\leq\frac{-b\pm\sqrt{b^2-4c}}{2}}\leq\frac{|b|}{2}\,$$ 
For example, assuming $\,\,b>0\,\,$: $$(1)\,\,-\frac{b}{2}\leq\frac{-b+\sqrt{b^2-4c}}{2}\Longleftrightarrow\frac{\sqrt{b^2-4c}}{2}\geq 0$$$$(2)\,\,\frac{-b+\sqrt{b^2-4c}}{2}\leq \frac{b}{2}\Longleftrightarrow\frac{\sqrt{b^2-4c}}{2}\leq b\Longleftrightarrow b^2-4c\leq 4b^2$$
And as both these inequalities are clear we're done.
A: if we rewrite  quadratic  equation as $x^2+(b*x)/a +c/a$  ,then we  have  following
$x_1+x_2=-b/a$
$x_1*x_2=c/a$
now  $c/a>0$  always, 
also $b>0$
and $b/a>c/a$
so we have followings
1.a>0
then we have  $x_1$ and $x_2$   are both negative ,so both of them is greater  then $-b/a$
2.a<0
then   then   then both $x_1$ and $x_2$   are positive,so  both of them is less then  $b/a$
A: $F(\frac{-b}{a}).F(\frac{b}{a}) = c(\frac{2b^2}{a} + c) = \frac{2b^2c}{a} + c^2 > 8c^2 + c^2 (as\ b^2 > 4ac) = 9c^2 > 0$ (ac>0 means that c is not 0).
So F(x) changes sign from -b/a to b/a, this together with the fact that it is a continuous function means that it must be 0 at some point in the interval.
EDIT: This does not actually work as it stands, we want the product to be negative not positive. We need to use product of F at -b/2a with F at -b/a instead. Here is the proof:
$F(\frac{-b}{a}).F(\frac{-b}{2a}) = c(\frac{-b^2}{4a} + c) < \frac{-4ac.c}{4a}+c^2$ (as $b^2 > 4ac\ $ and also, $\frac{c}{4a} > 0$ as $ac > 0$) $=-c^2 + c^2 = 0$.
This shows that we have a root between -b/a and -b/2a, so the looser condition of a root between -b/a and b/a is also satisfied.
