 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  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 Comments: ( Back to comment list | You must be logged in to post comments.) 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 realx, y, and z satisfying (1),(2), and (3).`
` `

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