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Re: quadratic formula
Posted:
Oct 17, 2012 10:22 PM


readability On Wed, 17 Oct 2012, stoneboy wrote:
> I have a little problem that I have not been able to get to the form > required. I may need some help. > > here is the problem: > > if p(x) = ax^2+bx+c is a second degree polynomial, then prove > > p(x) = a(x+b/2a)^2  ((b^2  4ac)/2a)
Have you writtent he problem correctly? No proof is possible. Consider the case b = 0, c /= 0.
If ax^2 + c = p(x) = ax^2 + 4ac/2a = ax^2 + 2c, then c = 0.
> Here I know the roots of the equation for the quadratic equation is > x=(b+sqrt(b^24ac))/2a
Pleaseusespacesforreadibility!
> and the negative of the radicand and also > x=b/2a is the vertex >
> Which root is substituted to get the above form. Or in other words, > how do I determine whether I use the negative or the positive radicand > in the above equation to get the form shown above. I have simplified > it all known ways but cannot seem to get it to that form. Am I > barking up the wrong tree here? > > thanks > > s >



