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 .