# SOLUTION: Question is: What can you say about the solutions of this nonlinear system? y=4x^2+4 y=2x^2-4 (a) The system has no solution (b) The system has two real solutions (c) The s

Algebra ->  Algebra  -> College  -> Linear Algebra -> SOLUTION: Question is: What can you say about the solutions of this nonlinear system? y=4x^2+4 y=2x^2-4 (a) The system has no solution (b) The system has two real solutions (c) The s      Log On

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 Question 30246: Question is: What can you say about the solutions of this nonlinear system? y=4x^2+4 y=2x^2-4 (a) The system has no solution (b) The system has two real solutions (c) The system has two imaginary solutions (d) The system has one solution of multiplicity 2 Please be specific on which is the correct answer- thank youAnswer by sdmmadam@yahoo.com(530)   (Show Source): You can put this solution on YOUR website!y=4x^2+4----(1) y=2x^2-4----(2) (1)-(2) (y-y) = (4x^2-2x^2)+[4-(-4)] 0 = 2x^2+8 dividing by 2 0 = x^2+4 (zero divided by anything(nonzero)is zero) That is x^2 +4 = 0 x^2 = -4 x= +2i or x=-2i which implies y =-12 Since there are two values to x and both imaginary, the result is the choice (c) The system has two imaginary solutions