AP EAMCET Engineering 2014 Code A Solved Paper

Consider 2 and 3 as they are smallest possible prime numbers (2 being the smallest).
Hence Case I
Let p=3 and q=2
x2−3x+2=0 implies
(x−2)(x−1)=0
x=2 and x=1
The roots are positive. Therefore 1,2 can be possible roots.
Case II p=2 and q=3
x2−2x+3=0
B2−4AC
=4−12
=−8
Now D<0.
Hence no real roots.
Therefore possible solutions are 1,2 .