algebra precalculus - What are the roots of this equation?

I have a quadratic equation $ax^2 +bx+c =0$, where $a,b,c$ all are positives and are in Arithmetic Progression.

Also, the roots $\alpha$ and $\beta$ are integers.

I need to find out $\alpha + \beta + \alpha\beta$.

I have tried taking $a = a' - d , b = a' , c = a' + d$ because I have supposed $a'$ is the first term and $d$ is common difference, I used sum of roots and product of roots rule, but nothing helped..

Answer

We see that the equation $$ax^2+(a+d)x+a+2d=0$$
has integer roots, which says $1+\frac{d}{a}\in\mathbb Z$.

Let $\frac{d}{a}=k$.

Hence, we have $$x^2+(1+k)x+1+2k=0$$
has integer roots, which gives

$$(1+k)^2-4(1+2k)=n^2,$$ where $n$ is non-negative integer number,
which gives
$$k^2-6k-3=n^2$$ or
$$(k-3)^2=n^2+12$$ or
$$(k-3-n)(k-3+n)=12$$ and the rest is smooth:

Since $n$ is non-negative integer, we obtain two cases only.

1. $n-k+3=6$ and $n+k-3=-2$, which gives $k=-1$;





2. $n-k+3=-2$ and $n+k-3=6$, which gives $k=7$.