Let a, b, c be positive integers in arithmetic progression such that the equation ax2+bx+c=0 has only integer solutions. Then which of the following statements is (are) TRUE ?
- c-b is an integer multiple of a
- Both the roots of the equation ax2+bx+c=0 are odd integers
- If c=15, then ab=8
- If b=8, then x=3 is a root of the equation ax2+bx+c=0
Answer
(1) c-b is an integer multiple of a, (2) Both the roots of the equation ax2+bx+c=0 are odd integers, (3) If c=15, then ab=8
Solution
Integer roots r,s: r+s=- ba, rs= ca are integers, so b=ka, c=la with k,l N; hence c-b=(l-k)a. (A) true.
AP: 2b=a+c l=2k-1. Equation becomes x2+kx+(2k-1)=0.
Discriminant k2-8k+4=(k-4)2-12 must be a perfect square m2: (k-4-m)(k-4+m)=12 with factors of equal parity \2,6\ or \-6,-2\ k=8 (since k>0).
So l=15: x2+8x+15=0 x=-3,-5, both odd. (B) true.
c=15 15a=15 a=1, b=8 ab=8. (C) true.
b=8 a=1, roots -3,-5; x=3 is not a root. (D) false.
Answer: (A), (B), (C)