Does a quadratic necessarily have a root in this interval? If F(x) is the quadratic ax2+bx+c with ac&gt;0 b2−4ac&gt;0, it is true that within the interval [−ba, +ba] there exists a point x where F(x)=0

Question

Does a quadratic necessarily have a root in this interval?  If F(x) is the quadratic ax2+bx+c with ac&amp;gt;0  b2−4ac&amp;gt;0, it is true that within the interval [−ba, +ba] there exists a point x where F(x)=0

votaren10 · Accepted Answer

You can do what J.M. suggested, i.e. use the quadratic formula. First, it&#039;s clear both a,c have the same sign, so multiplying by −1a if necessary we can assume the equation is x2+bx+c=0,c&amp;gt;0.Using the quad. form we get that we must prove−|b|2≤−b±b2−4c2≤|b|2For example, assuming b&amp;gt;0:(1)−b2≤−b+b2−4c2⇔b2−4c2≥0(2)−b+b2−4c2≤b2⇔b2−4c2≤b⇔b2−4c≤4b2And as both these inequalities are clear we&#039;re done.

coolbananas03ok · Accepted Answer

F(−ba).F(ab)=c(2b2a+c)=2b2ca+c2&amp;gt;8c2(ac&amp;gt;⊕c2(as b2&amp;gt;4ac)=9c2&amp;gt;0. means that c is not o).  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(−ba).F(−b2a)=c(−b24a+c)&amp;lt;4ac.c4a+c2(as b2&amp;gt;4ac and also, c4a&amp;gt;0 as ac&amp;gt;0)=−c2+c2=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.

RizerMix · Accepted Answer

if we rewrite quadratic equation as x2+(b⋅x)/a+c/a, then we have following x1+x2=−b/a x1⋅x2=c/a now c/a&amp;gt;0 always, also b&amp;gt;0 and b/a&amp;gt;c/a so we have followings 1. a&amp;gt;0 then we have x1 and x2 are both negative ,so both of them is greater then −b/a 2. a&amp;lt;0 then then then both x1 andx2 are positive,so both of them is less then b/a.

Does a quadratic necessarily have a root in this interval? If

Answered question

Answer & Explanation

New Questions in Algebra I