Given the following system #x^2+y^2-z^2+w^2=2 ; x+2y+z^2-3w^2=1# how do I show that if the couple of solution #(x_0,y_0,z_0,w_0)# satisfy #z_0*w_0 != 0# I can find solution for #z# and #w# depending to #x# and #y# ?

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Answer 1 · 2015-12-17T21:47:58

if #zw ne 0#, then #(x,y) notin {(-1, 1), (7/5,-1/5)}#

For eliminating z, #2 + 1 = x^2 + y^2 + w^2 + x + 2y - 3w^2#

#2w^2 = x^2 + y^2 + x + 2y - 3#

For eliminating w, #3 * 2 + 1 = 3x^2 + 3y^2 - 3z^2 + x + 2y + z^2#

#2z^2 = 3x^2 + 3y^2 + x + 2y - 7#

So, we got #((w^2),(z^2)) = 1/2 ((f_1(x,y)), (f_2(x,y)))#

And you don't want that #f_1 = f_2 = 0#

For eliminating #x^2#,

#3f_1 - f_2 = 0 Rightarrow 3x - x + 6y - 2y - 9 + 7 = 0#

#2x + 4y = 2 Rightarrow x = 1 - 2y#

#f_1 = 0 Rightarrow (1 - 2y)^2 + y^2 + 1 - 2y + 2y - 3 = 0#

#(1 - 4y + 4y^2) + y^2 - 2 = 0#

#5y^2 - 4y - 1 = 0#

#Delta = 16 + 4 * 5 = 36#

If Delta were < 0, there would be no (x,y) such that #f_1 = f_2 = 0#.

#y = (4 ± 6)/10 in {1,-1/5}#

#x = 1 - 2y in {-1, 7/5}#