Are you sure this is a GMAT problem. This problem need the concept of determinent.

d = b^2 - 4 ac

if b^2 - 4 ac > 0, then it intersects x axis twice.
if b^2 - 4 ac = 0, then x = -b/2a, so it intersects x axis once.
if b^2 - 4 ac <0 , it will not touch x axis. So no solution.

By now way we can check the second condition as we dont know the value of a,b and c.

1. a>0

b^2 - 4ac can be greater or less than 0, depending on the value of c.insufficient

2. c < 0

b^2 - 4ac can be greater or less than 0, depending on the value of a.insufficient

combinig both :

b^2 = positive, a>0 and c<0

so the equation is b^2 - 4a(-c) = b^2 + 4ac , so b^2 + 4ac > 0 ( both a and c are positive)

hence sufficient.

IMO:C

hope this helps !!

A touch easier way to think of this:

'a' will determine whether or not the parabola opens upward or downward.

'b' is meaningless to this problem because it shifts the parabola left and right

'c' This determines the vertex of the parabola --> the value of the 'y-coordinate'

(1) If a is negative then it is a parabola that faces downward. --> NS
(2) Determines vertex --> NS

Together they are sufficient because the parabola opens upward with a negative vertex, meaning it crosses the x-axis twice.

Keep these problems simple!