Q

[RIJU]

x^2 +2qx+ r =0 , ask for the numbers of intersection with X-axis.
1) q^2>r
2) r^2>q

[TwinnSplitter]

There's no way you can find the points of intersection without being given explicit values of q and r. Thus, E.

[arjmen]

I reckon the question has been framed a tad poorly.
It should have read, F(x) = x^2+2qx+r, number of intersections with the X-axis = ?

The curve will intersect the X-axis when F(x)=0, i.e. when x^2+2qx+r = 0
It is evident that if the above quadratic equation has
1. a pair of equal real roots, it will touch the x-axis at 1 point
2. a pair of complex or complex conjugate roots, it will never touch the x-axis
3. 2 distinct real roots, it will intersect the x-axis at 2 points

Determinant = 4*q^2 - 4r = 4*(q^2 - r)
So, to have distinct real roots, need q^2 > r

Statement !: Sufficient

Statement II: Useless data. Insufficient

answer A

[lego2401]

Agreed with the above solution

as long as b^2-4ac is positive, we have two real root. With two real roots,
In this case, b=2q and c = r and a=1
so we have 4q^2-4r
in st 1- since q^2 > r => deter. is positive
from the quadratic equation,
x = [-b +or- root(b^2-4ac)]/2a and this will give us two distinct values, so two x intercepts.
st 2- doesnt tell us how the determinant will be, and hence the number of roots.

(PS: Arjmen, sorry for the redundancy, just wanted to back the solution in more GMAT terms:: ). Not everyone here knows what complex roots, etc are, so could be intimidating to some.

[arjmen]

No worries Lego,
sometimes the engineer in me takes over