This is not a legitimate GMAT question. The GMAT does not test your knowledge of the equations of circles and parabolas in the way that an Algebra 2 or pre-Calculus class might.

Having said that, the way to solve:
Solve the second equation for x^2 ... x^2 = y+3
Plug that into the first equation: y + 3 + y^2 = 4
y^2 + y - 1 = 0

If you use the quadratic formula (which you never really need to do on the GMAT) you will get 2 solutions for y. (-1 +- root(5))/2

Each of those 2 solutions for y will have 2 corresponding x values, one positive and the other negative, for a total of 4 solutions.

Hi,

From the interface it looks like powerprep.


For the question at hand.

y = (x + root 3)(x - root 3)

1>draw a circle with radius = 2

2>Draw a parabola which cuts x axis at root 3 and -root 3(slightly less than 2). This parabola will cut the Y axis at -3 (put x =0 in given equation).



therefore if u draw properly, you will see that the cicle and parabola intersect at 4 points.

It does look like PowerPrep, but I still believe this question would not currently appear on the GMAT, since there are no questions in GMAT Prep or in the OG (that I am aware of) that test your knowledge of the equation of a circle in the coordinate plane.

Even questions about parabolas tend not to require much knowledge about the properties of the graph of a parabola (e.g. how to find the vertex, etc.). Of course, you should be able to find the solutions to an equation like y = x^2 - 3.

yup, agreed greg, but the OG does mention circles as within syllabus, i dont know whether equation of a circle is ever tested, however the basic equation's knowledge is too easy to be forgotten.