xyz radicals (Posted on 2008-10-13)

     3 = √x + √y + √z,
     3 = x√x + y√y + z√z,
     3 = x2√x + y2√y + z2√z.

	Submitted by Bractals
Rating: 4.0000 (1 votes)
Solution:	(Hide)
Adding the first and third equations and subtracting twice the second equation gives 3 + 3 - 2(3) = (1 + x2 - 2x)√x + (1 + y2 - 2y)√y + (1 + z2 - 2z)√z or (x - 1)2√x + (y - 1)2√y + (z - 1)2√z = 0 Each of the three terms is nonnegative. Therefore, (x - 1)2√x = (y - 1)2√y = (z - 1)2√z = 0 Therefore, x, y, and z must be 0 or 1. (1,1,1) is the only one, of the eight possible, that satisfies all three equations.

	Subject	Author	Date
	Answer	K Sengupta	2009-01-02 16:05:37
	Solution	Praneeth	2008-10-13 19:42:44
	Solution	Jonathan Lindgren	2008-10-13 15:57:59
	a humble start	Ady TZIDON	2008-10-13 12:14:16