
Re: quadratic formula
Posted:
Oct 17, 2012 9:46 PM


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...

