Question 30246
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