Three surfaces, one real point? (Posted on 2007-02-07)

	Submitted by K Sengupta
Rating: 5.0000 (1 votes)
Solution:	(Hide)
By the problem: (1)x^2 + 4yz+2z = 0; (2) x+2xy+ 2z^2 = 0; (3) 2xz+y^2+y+1 = 0 If x=0 then from (3), we obtain: y^2+y+1 = 0; which does not contain any real solution and is thus a contradiction. On the other hand, if z = 0, then from (1), we obtain: x = 0, which is not feasible for obvious reasons. Accordingly xz must be non-zero. From (1) and (2), we obtain: x^2 = -2z(2y+1) and 2z^2 = -x(2y+1); so that 2(x^2)(z^2) = 2xz(2y+1)^2 Or, xz = (2y+1)^2 Substituting this in (1), we obtain: 2(2y+1)^2 + y^2+y+1 = 0 Or, 3y^2 + 3y +1 = 0, which do not possess any real solution in y. Consequently, no solution is possible for the given problem.

	Subject	Author	Date
	Solution	Bractals	2007-02-10 12:17:44
	Solution	hoodat	2007-02-07 22:24:15
	Solution	TamTam	2007-02-07 17:14:44