<s>[INST] Real values of $k$ such that $kx^2+(k+3)x+k-3=0$ has $2$ positive integer roots 
There is only one real values of $k$ for which the quadratic equation $kx^2+(k+3)x+k-3=0$ has $2$ positive integer roots. Then the product of these two solutions is

What i tried.
$$kx^2+(k+3)x+k-3=0$$
for $k=0,$ i am getting $x=1$(only one integer roots)
for $\displaystyle k\neq 0, x=\frac{-k\pm \sqrt{(k+3)^2-4k(k-3)}}{2k}$
for integer roots, $(k+3)^2-4k(k-3)=p^2$
How do i solve it , Help me please 
 [/INST] There is no such $k.$

Since the product we're looking for is $$\frac {k-3}{k},$$ first we must have $k\ne 0.$ Also, since the roots are positive integers, it follows that their product is also a positive integer. Thus, $k\ne 3$ and we also have that $k-3$ and $k$ must have the same sign.
Having noticed this, we find that the discriminant $$(k+3)^2-4k(k-3)$$ must be a nonzero square. Thus we must have $$0<(k-3)^2<12.$$ It follows that $(k-3)^2$ is either $1,4,$ or $9.$ Thus, $k-3$ must be some of $\pm 1,\pm 2, \pm 3,$ so that correspondingly, $k$ is $4,2;5,1;6,0.$
Now using the fact that $k$ is nonzero and has the same sign as $k-3$ leaves just $k$ being one of $4,5,6.$ Finally using the condition that $(k-3)/k$ be integral leaves us no such value of $k.$
Thus, the claim that there exists such a $k$ is false. Your conditions are too restrictive. Completely killed off the $k$'s.
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