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

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 William Elliot Posts: 2,490 Registered: 1/8/12
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^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
>