Question

Suppose the quadratic polynomial $P(x) = ax^2 + bx+ c$ has positive coefficients $a, b, c$ in arithmetic progression in that order. If $P(x) = 0$ has integer roots $\alpha$ and $\beta$. Then, $\alpha + \beta + \alpha \beta$ is equal to

• A. 3
• B. 5
• C. 7
• D. 14

Answer 1

Answer: (C)

We have, $p(x) = ax^2 + bx + c$, where

$a, b, c$ are in $AP$ and $a, b, c$ are positive real.

$\alpha, \beta$ are root of $p(x) = 0$, where $\alpha$ and $\beta$ are integers.

$p(x) = ax^2 + bx + c = 0$

$\alpha + \beta = \frac{-b}{a}, \alpha \beta = \frac{c}{a}$

$\alpha, \beta$ are integer.

$\therefore \alpha + \beta = \frac{-b}{a} = -\lambda , \lambda \in I$

$\Rightarrow b = a\lambda$

$a, b, c$ are in $AP$.

$\therefore b = \frac{a + c}{2}$

$\Rightarrow \frac{a + c}{2} = a\lambda$

$\Rightarrow c = a ( 2\lambda - 1)$

$\therefore ax^2 + a\lambda x + a(2\lambda - 1) = 0$

$\Rightarrow x^2 + \lambda x + (2\lambda -1) = 0 [ \because a \ne 0]$

$D = \lambda^2 - 4 (2\lambda - 1)$ is a perfect square for integral roots.

$\therefore \lambda^2 - 8\lambda + 4 = k^2$

$\Rightarrow (\lambda - 4)^2 - 12 = k^2$

$\Rightarrow (\lambda - 4 - k)(\lambda - 4 + k) = 2\times 6$

$\Rightarrow \lambda - 4 - k = 2$ and $\lambda - 4 + k = 6$

$\because \lambda = 8 $ and $ k = 2$

$\therefore \alpha + \beta + \alpha \beta = \frac{-b}{a} + \frac{c}{a}$

$= \frac{-a\lambda + a(2\lambda - 1)}{a}$

$= \frac{a(\lambda - 1)}{a}$

$ = \lambda - 1 = 8 - 1 = 7$