Quadratic equation - arithmetic progression

Question

The coefficients of a quadratic equation $ax^2+bx+c=0$ are in arithmetic progression and $a, b, c$ all are positive. If the roots of the equation be $k$ and $l$ are integers.Then $k +l+kl$ =?

Answer 1

By Vieta's Theorem, we get that $k+l=-\frac{b}{a}$ and $kl=\frac{c}{a}$.

Also, $a+c=2b$ as $a,b,c$ are in A.P.

So $\boxed{k+l+kl=-\frac{b}{a}+\frac{c}{a}=\frac{c-b}{a}=\frac{c-b}{2b-c}}$

Answer 2

Let $b-a=c-b=d\implies ax^2+(a+d)x+a+2d=0$

Now $k+l+kl=(k+1)(l+1)-1$

Now if $y=x+1\iff x=y-1$

$0=a(y-1)^2+b(y-1)+c=ay^2+y(\cdots)+a-b+c=0$

Now as $a+c=2b$

$\implies(k+1)(l+1)=\dfrac{a-b+c}a=\dfrac ba$