kx² + 20x + 20 - k = 0

If there are to be only integer solutions, then the sum of the roots and the product of roots must also be integers.


Sum of roots (-'b/a') = -20/k

Product of roots ('c/a')
= (20 - k)/k
= 20/k - 1

So, as long as 20/k is an integer, it will satisfy the conditions.

As long as k is a factor of 20, 20/k will be an integer.

Factors of 20 are : 1,2,4,5,10,20

So we have 6 possible integer values of k here.

Their negatives will work as well. So 6 × 2 = 12


Now if we consider k = 0, the equation becomes a linear one.

20x + 20 = 0
x + 1 = 0
x = -1


So in total, there can be 13 possible integer values of k.

But, since the question mentions quadratic, then we should be able to ignore the case where k = 0.

So 12 possible integer values of k