The Math Forum

Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Math Forum » Discussions » Math Topics » alt.math.undergrad

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: quadratic formula
Replies: 5   Last Post: Oct 21, 2012 11:06 AM

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ] Topics: [ Previous | Next ]
William Elliot

Posts: 2,637
Registered: 1/8/12
Re: quadratic formula
Posted: Oct 17, 2012 10:22 PM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

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^2-4ac))/2a


> 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

Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© The Math Forum at NCTM 1994-2018. All Rights Reserved.