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April 17, 2014

April 17, 2014

Posted by **Greg B.** on Tuesday, June 12, 2012 at 3:50pm.

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!

- Quadratic Equations -
**Steve**, Wednesday, June 13, 2012 at 10:39amThe 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 downward-opening 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.

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