When k is an integer, the equation (K & # 178; - 1) x & # 178; - 6 (3K-1) x + 72 = 0 has two unequal positive integer roots? Good bonus points to write, the process is urgent!

When k is an integer, the equation (K & # 178; - 1) x & # 178; - 6 (3K-1) x + 72 = 0 has two unequal positive integer roots? Good bonus points to write, the process is urgent!

Cross multiplication
[(k-1)x-6][(k+1)x-12]=0,
x1=6/(k-1),x2=12/(k+1)
k-1=1,2,3,6
k=2.3.4,7
k+1=1,2,3,4,6,12
k=0,1,2,3,5,11
The common K has 2,3
The two are not equal
When k = 3, both are 3
So k = 2

When k takes what value, the equation (k2-1) x2-6 (3K-1) x + 72 = 0 has two unequal positive integer roots?
I can't type the square. You should be able to understand it

First of all, by using cross multiplication, we can have [(k-1) X-6] [(K + 1) X-12] = 0,
Then we can get two roots x = 6 / (k-1) and 12 / (K + 1) respectively
Then, because it is a positive integer, if 6 / (k-1) is a positive integer, K can take 7,4,3,2
Considering that 12 / (K + 1) is also a positive integer, we substitute four values into the checking calculation and find that 3 and 2 satisfy the condition of positive integer. However, when k = 3, the two roots are 3 at the same time, which is not equal, so K can not be equal to 3
Then k = 2

When k is an integer, the equation (k ^ 2-1) x ^ 2-6 (3K-1) x + 72 = 0. There are two unequal positive integer roots

K ^ 2-1 is not equal to 0
【6（3k-1）】^2-72*4(k^2-1)=36(9k^2-6k+1)-288k^2+288>0
Just solve it!