$$ax^{2} + (a+3)x + a - 3=0$$ This is the quadratic equation, and the question asks for product of roots. (a) $9$ (b) $8$ (c) $6$
This is how I proceeded. As far as I know if roots are integers then the Discriminant is going to be perfect square. So, $D=(a+3)^2 - 4(a-3)a\qquad D+3(a - 3)^2= 36$
So, is there any way I can find the value of $a$. Or this approach is totally wrong... Thanks for help
The coefficients/roots formula tells you that the product of the roots is going to be $\frac{a-3}{a}$. If this is an integer, say $n$, then $a=\frac{3}{1-n}$ is rational.
From there, your argument that $D$ has to be a perfect square is valid. However, it doesn't seem obvious how to list the values taken by a polynomial function that are perfect squares.
An easier way from here is to use the hint: $n\in\{9,8,6\}$. Therefore $a\in \{-\frac38, -\frac37, -\frac35\}$ and it boils down to hand checking.
Another approach is to use both coefficients/roots formulas:
$x_1+x_2=\frac{-a-3}{a}$ and $x_1x_2=\frac{a-3}{a}$ imply $$x_1x_2=x_1+x_2+2$$
This equation has a very limited number of solutions that are integers, since the left-hand side grows much faster than the right-hand side.