Question

Let $a, b, c$ be positive integers in arithmetic progression such that the equation \[ ax^2 + bx + c = 0 \] 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 $ax^2 + bx + c = 0$ are odd integers
• C. If $c = 15$, then $ab = 8$
• D. If $b = 8$, then $x = 3$ is a root of the equation $ax^2 + bx + c = 0$

Hint:

When given that roots are integers, always try to use the product and sum relations to create a factorable algebraic expression. Adding a constant to both sides to complete the factorization (Simon's Favorite Factoring Trick) is a very powerful technique in such integer-solution problems.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The problem involves a quadratic equation with positive integer coefficients $a, b,$ and $c$ that form an arithmetic progression (AP).
The condition that the equation has only integer solutions implies that the roots must be integers.
We need to determine the specific values or relationships between these coefficients using the properties of roots and AP.

Step 2: Key Formula or Approach:

• For $a, b, c$ in AP, we have the condition $2b = a + c$.

• Let the integer roots of the equation be $\alpha$ and $\beta$.

• From Vieta's formulas: $\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.

• Since $a, b, c$ are positive integers, the sum of roots $\alpha + \beta$ must be negative and the product $\alpha\beta$ must be positive. This implies both roots $\alpha$ and $\beta$ must be negative integers.

Step 3: Detailed Explanation:

• From the AP condition $2b = a + c$, we can divide by $a$ to get $2\left(\frac{b}{a}\right) = 1 + \frac{c}{a}$.

• Substituting the expressions from Vieta's formulas: $2(-\alpha - \beta) = 1 + \alpha\beta$.

• Rearranging the terms, we get: $\alpha\beta + 2\alpha + 2\beta + 1 = 0$.

• To factorize this expression, we add 3 to both sides: $\alpha\beta + 2\alpha + 2\beta + 4 = 3$.

• This factors as $(\alpha + 2)(\beta + 2) = 3$.

• Since $\alpha$ and $\beta$ are integers, $(\alpha + 2)$ and $(\beta + 2)$ must be integer factors of 3.

• Possible cases for $(\alpha + 2, \beta + 2)$ are $(1, 3)$, $(-1, -3)$, $(3, 1)$, or $(-3, -1)$.

• Case 1: $\alpha + 2 = 1$ and $\beta + 2 = 3 \implies \alpha = -1, \beta = 1$. Here $\alpha + \beta = 0$, so $-\frac{b}{a} = 0 \implies b = 0$. This contradicts that $b$ is a positive integer.

• Case 2: $\alpha + 2 = -1$ and $\beta + 2 = -3 \implies \alpha = -3, \beta = -5$.

• For these roots, $\alpha + \beta = -8 \implies -\frac{b}{a} = -8 \implies b = 8a$.

• Also, $\alpha\beta = 15 \implies \frac{c}{a} = 15 \implies c = 15a$.

• Verification of AP: $2b = 16a$ and $a + c = a + 15a = 16a$. The condition holds.

• Checking (A): $c - b = 15a - 8a = 7a$. Since $a$ is an integer, $c - b$ is an integer multiple of $a$. Statement (A) is TRUE.

• Checking (B): The roots are $-3$ and $-5$. Both are odd integers. Statement (B) is TRUE.

• Checking (C): If $c = 15$, then $15a = 15 \implies a = 1$. Then $b = 8(1) = 8$. Thus $ab = 1 \times 8 = 8$. Statement (C) is TRUE.

• Checking (D): If $b = 8$, then $8a = 8 \implies a = 1$. The roots are $-3$ and $-5$. Thus $x = 3$ is not a root. Statement (D) is FALSE.

Step 4: Final Answer:
The true statements are (A), (B), and (C).