Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
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

3 Answers 3

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.

share|improve this answer
    
+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? –  flapjackery 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.

share|improve this answer

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 $b/a$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.