<s>[INST] 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..
 [/INST] If the roots are integers, we can write $x^2 + bx + c = (x-\alpha)(x-\beta)$. Note that it's okay to take $a=1$, since $b-a = c-b \implies \frac{b-a}{a} = \frac{c-b}{a}$, i.e. the arithmetic progression is preserved.
Now write $c = b + (b-1) = 2b-1$.
Use Vieta's: $\alpha + \beta = -b$ and $\alpha\beta  = 2b-1$.
$\alpha = \frac{-b\pm \sqrt{b^2 - 8b +4}}{2}$. But $\alpha$ is an integer, so $(b-4)^2 -12$ is a perfect square means that we must have $b = 8$ (can be checked manually by trying squares up to 6, since difference between consecutive squares is $2n+1$). Then $\alpha + \beta + \alpha\beta = -b + (2b-1) = b - 1 = 7$.
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