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Re: quadratic formula
Posted:
Oct 17, 2012 9:46 PM
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On Wed, 17 Oct 2012 20:14:05 -0500, stoneboy wrote:
>
> Hi,
>
> 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)
= a(x^2 + 2xb/2a + b^2/4a^2) - b^2/2a + 2c
= ax^2 + bx + b^2/4a - b^2/2a + 2c
= ax^2 + bx - b^2/4a + 2c
So either -b^2/4a + 2c = c, which means b^2/4a must equal c (which
doesn't seem right), or you've transcribed the formula incorrectly or
left out some condition, or I've made a mistake in my algebra.
--
Stan Brown, Oak Road Systems, Tompkins County, New York, USA
http://OakRoadSystems.com
Shikata ga nai...
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