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.
Does a quadratic necessarily have a root in this interval?
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0@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 – 2012-05-14
3 Answers
$F(\frac{-b}{a}).F(\frac{b}{a}) = c(\frac{2b^2}{a} + c) = \frac{2b^2c}{a} + c^2 > 8c^2 + c^2 (as\ b^2 > 4ac) = 9c^2 > 0$ (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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0Actually 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. – 2012-05-14
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.
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 $b/a$