JEE Advanced 2026 — Paper 2 — MATH-II solution

JEE Advanced2026Paper 2MATH-II

Let a, b, c be positive integers in arithmetic progression such that the equation

$a x^{2} + b x + c = 0$

has only integer solutions.

Then which of the following statements is (are) TRUE?

(A) $c - b$ is an integer multiple of $a$

(B) Both the roots of the equation $a x^{2} + b x + c = 0$ are odd integers

(D) If $b = 8$ , then $x = 3$ is a root of the equation $a x^{2} + b x + c = 0$

A
$c - b$ is an integer multiple of $a$
B
Both the roots of the equation $a x^{2} + b x + c = 0$ are odd integers
C
If $c = 15$ , then $ab = 8$
D
If $b = 8$ , then $x = 3$ is a root of the equation $a x^{2} + b x + c = 0$

Solution

Since $a, b, c$ are positive integers in AP, $2 b = a + c$ . Let the integer roots be $α, β$ . Then $a x^{2} + b x + c = a (x - α) (x - β)$ .

Evaluate at $x = - 2$ : $4 a - 2 b + c = a (- 2 - α) (- 2 - β) = a (α + 2) (β + 2)$ . Using $2 b = a + c$ , the left side becomes $4 a - (a + c) + c = 3 a$ . Hence $(α + 2) (β + 2) = 3$ .

Since $a, b, c > 0$ , both roots are negative, so $α + 2, β + 2$ are integers whose product is $3$ and at most $2$ . The only factorizations give $(α + 2, β + 2) \in {(- 1, - 3), (- 3, - 1)}$ , so ${α, β} = {- 3, - 5}$ .

From Vieta: $α + β = - \frac{b}{a} = - 8$ so $b = 8 a$ ; $α β = \frac{c}{a} = 15$ so $c = 15 a$ .

(A): $c - b = 15 a - 8 a = 7 a$ , a multiple of $a$ . Correct.

(B): Both roots are $- 3$ and $- 5$ , which are odd integers. Correct.

(C): If $c = 15$ , then $a = 1$ and $b = 8$ , so $ab = 8$ . Correct.

(D): If $b = 8$ , then $a = 1, c = 15$ . The roots are $- 3$ and $- 5$ , not $3$ . Incorrect.

$∴$ Correct options: (A), (B), (C).

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