parabola will intersect in two points x- axis if
p^2-4r>0, or p^2>4r
it will only touch x-axis if p^2=4r
and it will not intersect x-axis if p^2-4r<0 or p^2<4r
at any rate we need to compare p^2 and 4r
(1)p^2>r to me not suff,
p=2, r=1, 2^2>1, but p^2=4 will equal 4r=4*1. thus only one point
or p=3 r=1, 9>1 and 9>4*1 here we see intersection in two points
(2)r^2>p, again insuff
r=-2 p=1, (-2)^2>1 and p^2=1>4*(-2) i mean p^2>4r, thus two points intersection
p=-2, r=1. r^2=1>-2, but p^2=(-2)^2=4 and 4r=4*1, 4=4 intersection in one point
both are too hard for me
honestly no idea how to solve
p^2>r and r^2>p in a while
my answer is E

Thanks Man! nice job attempting...the problem scared me to the extent I forgot the b^2-4ac concept.

However this reminds. Let me attempt the rest. Atleast you arrived at the right answer.

Together, p^2>r and r^2>p prolly no way to reduce it to a single equation without jeopardizing the signs, so the best way is to substitute values
p=2,r=-3; 4>-3 and 9>2 but (4+4.-3) <0 no roots. However,
p=5,r=3; 25>3 and 9>5 and 25-12>0 so two roots. INSUFF