Determine whether there exist real numbers x, y and z satisfying the following system of equations:

x²+4yz+2z = 0; x+2xy+2z² = 0; 2xz+y²+y+1 = 0

There is no solution to this problem.

Starting with the third equation, if either x=0 OR z=0, then the resulting equation is [y²+y+1 = 0].  This results in complex roots for y.  Therefore, neither x nor z can be zero.  We also know that the preceding equation for y must be greater than one (there are no real numbers that will produce a negative result for (y²+y+1).  Thus, either (x<0 AND z>0) OR (x>0 AND z<0).

Rewrite equation 1 as [x²=-2z(2y+1)].  Since x²>0, either (z<0 AND y>-0.5) OR (z>0 AND y<-0.5) to make the right side of the equation positive.

Rewrite equation 2 as [2z² = -x(2y+1)].  Since z²>0, either (x>0 AND y<-0.5) OR (x<0 AND y>-0.5).

This leaves these contradicting conditions:

Case 1:  y>-0.5   ----->      z<0 AND x<0

Case 2:  y<-0.5   ----->      z>0 AND x>0

This makes both x AND z positive or both x AND z negative.  However, this is not possible in equation 3.

Posted by hoodat on 2007-02-07 22:24:15