Product of roots of $ax^2 + (a+3)x + a-3 = 0$ when these are positive integers There is only one real value of $'a'$ for which the quadratic equation $$ax^2 + (a+3)x + a-3 = 0$$ has two positive integral solutions.The product of these two solutions is : 
Since the solutions are positive, therefore the product of roots and sum of root will be positive. This will give us two inequalities in $a$. Substitute the values of $a$ in the quadratic but I'm not getting my answer correct.
The two inequalities will be $$\frac{a-3}{a} > 0$$ and the other one will be $$\frac {a+3}{a} < 0$$
From the first one we get $a>3$ and from the second one we get $a< -3$.
I don't know how to proceed after this.
Kindly help.

We need $$\frac{a+3}{a}\in\Bbb{Z}\ ,  \frac{a-3}{a}\in\Bbb{Z}$$ or $$1+\frac{3}{a}\in\Bbb{Z}\ , \ 1-\frac{3}{a}\in\Bbb{Z}$$ thus $\displaystyle \frac{3}{a}\in\Bbb{Z}$, means that $\displaystyle a=\frac{3}{m}$ where $m\in\Bbb{Z}$.
Now we can write the equation as $$\frac{3}{m}x^2+\left(\frac{3}{m}+3\right)x+\frac{3}{m}-3=0\implies x^2+(m+1)x+1-m=0$$
Using quadratic formula we have $$x_{1,2}=\frac{-m-1\pm\sqrt{m^2+6m-3}}{2}=\frac{-m-1\pm\sqrt{(m+3)^2-12}}{2}$$We need the expression under the square root to be an integer. There are only two squares with difference $12$ (those are $4,16$), hence we want $(m+3)^2=16\implies m=1\text{ or }m=-7$, thus $$a=3 \text{ or } a=-\frac{3}{7}$$ and as $a<0$ we have 

$$a=-\frac{3}{7}$$

