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. $3$
• B. $5$
• C. $7$
• D. $14$

Answer 1

Answer: (C)

$7$

If $a,b,c$ are in arithmetic progression, then $2b=a+c$

$\alpha +\beta +\alpha \beta =$ (sum of roots) + (product of roots)

$\Rightarrow \alpha +\beta +\alpha \beta =-\frac{b}{a}+\frac{c}{a}=\frac{c-b}{a}=\frac{b-a}{a}=\frac{b}{a}-1$

$\alpha +\beta +\alpha \beta$ is an integer $\Rightarrow a|b$ ($a$ divides $b$)

$2b=a+c\Rightarrow a|c$

Write $(a,b,c)$ as $(a,a(1+r),a(1+2r\left)\right)$

Now, determinant of $p\left(x\right)$ should be a square since roots are integers.

$D=b^2-4ac=a^2(1+r{)}^2-4a\cdot a(1+2r)=a^2[(1+r{)}^2-4(1+2r\left)\right]$

$\Rightarrow (1+r{)}^2-4(1+2r)=k^2$
$\Rightarrow r^2-6r-(3+k^2)=0$

This is a new quadratic equation in $r$. $r$ is also an integer, hence its determinant is also a square.

$D^{\prime }=36-4[-(3+k^2\left)\right]=4(12+k^2)$

$\Rightarrow 12+k^2=p^2$
$\Rightarrow (p-k)(p+k)=12$

Looking for integral solutions,

$(k,p)\in \left\{\right(2,4),(-2,-4),(2,-4),(-2,4\left)\right\}$

$\Rightarrow k=\pm 2$
$\Rightarrow k^2=4$

$r=7$ or $r=-1$

For $r=-1$, $D$ is negative. Not possible.

For $r=7$ , the equation is $a{x}^2+8ax+15a=0$. Its roots are $-3$ and $-5$.

Hence, $\alpha +\beta +\alpha \beta =7$