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 ->  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


   

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 you
Answer by sdmmadam@yahoo.com(530) About Me  (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