Question

Suppose the quadratic polynomial $P\left(x\right)=a{x}^2+bx+c$ has positive coefficients $a,b,c$ in arithmetic progression in that order. If $P\left(x\right)=0$ has integer roots $\alpha$ and $\beta$, then $\alpha +\beta +\alpha \beta$ equals

• A. $14$
• B. $5$
• C. $3$
• D. $7$

Answer 1

The correct option is D $7$

$P\left(x\right)=a{x}^2+bx+c=0$ has integer roots $\alpha$ and $\beta$.

Taking, $\alpha +\beta +\alpha \beta =-\frac{b}{a}+\frac{c}{a}=\frac{c-b}{a}...\left(1\right)\text{Since}a,b\&c\text{are in A.P.},\Rightarrow a+c=2b\Rightarrow c-b=b-a\Rightarrow \alpha +\beta +\alpha \beta =\frac{b}{a}-1\text{Let}\frac{b}{a}=z\text{(an integer)}\Rightarrow b=az\text{and}c=2b-a=a(2z-1)$

$\therefore P\left(x\right)=a{x}^2+azx+a(2z-1)=a[x^2+zx+(2z-1\left)\right]$

$\text{Discriminant}\left(D\right)=z^2-8z+4,\text{should be a perfect square}\text{Let}D=(z-4{)}^2-12=m^2,m\in I\Rightarrow (z-4+m\left)\right(z-4-m)=12[12=12\times 1,6\times 2\text{or}4\times 3]$
But for $(12,1)$ and $(4,3)$, $z$ will not be an integer.
$\Rightarrow z-4+m=6\&z-4-m=2$
$\Rightarrow z=8\Rightarrow \alpha +\beta +\alpha \beta =\frac{b}{a}-1=8-1=7$