Question

If a,b are distinct prime numbers such that x²-ax+b=0 has positive integral root then (a+b) can be​

Answer 1

Answer:

Consider 2 and 3 as they are smallest possible prime numbers (2 being the smallest).

Hence Case I

Let p=3 and q=2

x

2

−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

x

2

−2x+3=0

B

2

−4AC

=4−12

=−8

Now D<0.

Hence no real roots.

Therefore possible solutions are 1,2.