Quadratic Equations
posted by Greg B. on .
The question is
Is the point (3, 2) a solution of the intersection of the following set of quadratic equations:
Y < X^2
X^2 + Y^2 < 16
I guess I am somewhat confused by the way it's written. Would I be graphing this to find out? I'm so confused with this problem. I have asked for help and alot of people really don't understand it when they look at it. The answer I have for it is 321, but I am not quite sure if that really answers the question. Is the point a solution of the intersection of those equations? Please help me, I feel so lost!

The solution to inequalities is a region, not just a point. In this case, think of the graphs.
y < x^2 is the whole inside of the downwardopening parabola with vertex at (0,0).
x^2 + y^2 < 16 is the inside of the circle of radius 4 centered at (0,0).
So, is 9+4<16? yes, the point is inside the circle.
Is 2 < 9? No. It is not inside the parabola.
How can you possibly come up with 321 as an answer? Is that x or y?
The point is not a solution of equations, because you don't have any equations; you have inequalities.
visit wolframalpha . com and enter
plot {x^2+y^2 < 16, y < x^2}
to see the shaded region which satisfies both inequalities.