If K is a positive integer and the equation (k ^ 2-1) x ^ 2-6 (3K-1) x + 72 = 0 about X has two unequal positive integer roots, find the value of K

If K is a positive integer and the equation (k ^ 2-1) x ^ 2-6 (3K-1) x + 72 = 0 about X has two unequal positive integer roots, find the value of K

First of all, because there are two real roots, we can infer that f (x) = (k ^ 2-1) x ^ 2-6 (3K-1) x + 72 is a quadratic function of one variable
Then we can get k ^ 2-1 ≠ 0, that is k ≠ ± 1
Secondly, because there are two unequal real roots, we can get that the discriminant is greater than 0,
That is [6 (3K-1)] ^ 2-4x (k ^ 2-1) X72 ＞ 0, the simplification result is (K-3) ^ 2 ＞ 0, which only needs K ≠ 3
Because they are two positive integer roots, the sum of the two and the product of the two should be greater than 0
That is, 6 (3K-1) / (k ^ 2-1) > 0, 72 / (k ^ 2-1) > 0
Finally, I'll simplify myself. After simplifying, I'll synthesize several conditions that limit the value of K. I'm just providing ideas