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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.

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You've tried using the quadratic formula here? – J. M. May 14 '12 at 11:10
@flapjackery you need to show that,whenever a is positive or negative,so is c,it means that roots have also same sign,now range depend on only what is possible values of -b/a,and after this, you can easily follow instructions,see my answer – dato datuashvili May 14 '12 at 12:10
up vote 2 down vote accepted

$F(\frac{-b}{a}).F(\frac{b}{a}) = c(\frac{2b^2}{a} + c) = \frac{2b^2c}{a} + c^2 > 8c^2 + c^2 (as\ b^2 > 4ac) = 9c^2 > 0$ (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.

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+1 @Wonder for nice explanation – dato datuashvili May 14 '12 at 12:33
@Wonder Great explanation. I see there must be a root in the interval. So would this also imply that the points at F(b/a) and F(-b/a) must have different signs themselves? Or only that a sign change must exist? – mathjacks May 14 '12 at 13:56
Actually if we look at the first solution I incorrectly gave, F(b/a) and F(-b/a) will in fact have the same sign. It will be different from the sign of F(-b/2a) though. – Wonder May 14 '12 at 17:34

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.

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if we rewrite quadratic equation as $x^2+(b*x)/a +c/a$ ,then we have following



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 $b/a$

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