When k takes what value, the square of the inequality 2kx + KX + 3 / 8 > 0 holds for all real numbers X

When k takes what value, the square of the inequality 2kx + KX + 3 / 8 > 0 holds for all real numbers X

2kx²+kx-3/8

It is known that the solution set of inequality kx-2 > 0 (k is not equal to 0) about X is X

Transfer of items
kx＞2
Known x ＜ - 1
Take it in. It's time
k＜-2
It should be right,

The inequality (kx-k * k-1) (X-2) about X is known to be > 0, where k is not equal to 1
(1) Try to write out the necessary and sufficient conditions for the inequality to be a linear inequality of one variable;
(2) Write the solution set a of the inequality;
(3) From (2), if the intersection of a and Z is B, try to explore whether the number of elements in the set B is limited? If so, find out the value range of K which makes the number of elements in the set B minimum, and write B with enumeration method; if not, please explain the reason

The inequality (kx-k * k-1) (X-2) about X is known to be > 0, where k is not equal to 1
(1) Try to write out the necessary and sufficient conditions for the inequality to be a linear inequality of one variable;
(kx-k*k-1)(x-2)>0
kx^2-x*k^2-x-2kx+2k^2+2>0
A necessary and sufficient condition for linear inequality of one variable, k = 0,
(2) Write the solution set a of the inequality;
-(x-2)>0
x-2