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Re: quadratic formula
Posted:
Oct 18, 2012 8:11 AM


my apologies. I should have used spaces for clarity. add to that I made a typo. darn it. sorry to waste your time folks.
if p(x) = ax^2 + bx + c is a second degree polynomial, then prove
p(x) = a(x+b/2a)^2  ( (b^2  4ac) / 4a )
funny how they want me to derive the latter from the former. Guess I am a little lost.
s
On Wed, 17 Oct 2012 19:22:35 0700, William Elliot <marsh@panix.com> wrote:
>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 >>



