Question

Suppose the quadratic polynomial $P(x)=a{x}^2+bx+c$ 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, $p(x)=a{x}^2+bx+c$, 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.

$p(x)=a{x}^2+bx+c=0$

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

$\alpha ,\beta$ are integer.

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

$\Rightarrow b=a\lambda$

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

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

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

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

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

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

$D={\lambda }^2-4(2\lambda -1)$ 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 $k=2$

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

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

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

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