SOLUTION: Can you help me solve this system? x^2+y^2=4 and y=x^2-2

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Question 387221: Can you help me solve this system? x^2+y^2=4 and y=x^2-2
Answer by robertb(4012) About Me  (Show Source):
You can put this solution on YOUR website!
+x%5E2%2By%5E2=4+ and y=x%5E2-2.
The 2nd equation becomes y+%2B+2=x%5E2. Substitute.
y%5E2+%2B+y+%2B+2+=+4, or y%5E2+%2B+y+%2B+-2+=+0.
(y+2)(y - 1) = 0.
y = -2, y = 1. When y = -2, then -2+%2B+2+=+x%5E2, so x = 0.
When y = 1, then 1%2B2+=+x%5E2, or x%5E2+=+3, or x = -sqrt%283%29, sqrt%283%29. Therefore there are 3 solutions: (-sqrt%283%29, 1), (sqrt%283%29, 1), and (0, -2). The solutions correspond to the 3 points of intersection of the parabola y=x%5E2-2 and the circle +x%5E2%2By%5E2=4+.