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Three surfaces, one real point? (Posted on 2007-02-07) Difficulty: 3 of 5
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

See The Solution Submitted by K Sengupta    
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Solution Solution Comment 3 of 3 |

 x^2+2z(1+2y) = 0                (1)
 2z^2+x(1+2y) = 0                (2)
 y^2+y+1+2xz  = 0                (3)
Multiply (1) by x to get
 x^3+2xz(1+2y) = 0               (4)
Multiply (2) by -2z to get
 -4z^3-2xz(1+2y) = 0             (5)
Adding (4) and (5) gives
 x^3 = 4z^3                      (6)
For x and z real, (6) implies
 xz >= 0                         (7)
Solving (3) for y gives
 y = (-1+-sqrt(1-4(1+2xz))/2     (8)
For y real, (8) implies
 xy < 0
This conflicts with (7).
Therefore, there are no real
x, y, and z satisfying (1),
(2), and (3).
 

  Posted by Bractals on 2007-02-10 12:17:44
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