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 Three surfaces, one real point? (Posted on 2007-02-07)
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 No Rating

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 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)
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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