3 Distinct Roots Not (Posted on 2008-03-04)

	Submitted by Bractals
Rating: 4.0000 (2 votes)
Solution:	(Hide)
Assume false. Let f(x) = x3 + 2px2 + 2p2x + p. Then there exists real numbers x1, x2, and x3 such that x1 < x2 < x3 and f(x1) = f(x2) = f(x3) = 0 Since f is differentiable everywhere, Rolle's theorem implies there exists real numbers y1 and y2 such that x1 < y1 < x2 < y2 < x3 and f'(y1) = f'(y2) = 0 But, f '(x) = 3x2 + 4px + 2p2 and (4p)2 - 4(3)(2p2) > 0 is a contradiction.

	Subject	Author	Date
	3 distinct solutions	Eigenray	2008-03-07 19:22:53
	re: Viete!	K Sengupta	2008-03-05 13:08:07
	Viete!	Chesca Ciprian	2008-03-05 07:43:13
	solution	Charlie	2008-03-04 11:13:44