The zeros of a quadratic function

"Determine the values of k for which the function f(x) = 4x squared - 3x +2kx +1 has two zeros."

"Calculate the value of k such that kx squared - 4x +k = 0 has one root."

Whoever answers this so I understand get's best answer!

f(x) = 4x^2 - 3x + 2kx + 1
f(x) = 4x^2 - (3 - 2k)x + 1
Find the discriminant. b^2 - 4ac
(3 - 2k)^2 - 4*4*1
(3 - 2k)^2 - 16
This must be greater than 0 to have two zeros.
(3 - 2k)^2 - 16 > 0
(3 - 2k)^2 > 16
3 - 2k > +√16
3 - 2k > 4
k < -1/2
3 - 2k < -√16
3 - 2k < -4
k > 7/2, k < -1/2

kx^2 - 4x + k = 0
4^2 - 4*k*k = 0
16 - 4k^2 = 0
k^2 = 4
k = 2, k = -2

1) ax^2 + bx + c = 0 has two roots or zeros, if
b^2 is more than 4ac, so by comparison,

4x^2 + (2k - 3)x +1 has two zeros if
(2k - 3)^2 is more than 4*4*1. Expand that.
4k^2 - 12k - 7 > 0
Compare the erect parabola
g(x) = 4x^2 - 12x - 7 = (2x + 1)(2x - 7)
Positive values for g and hence k occur when k < -1/2 or > +7/2

2) kx^2 - 4x + k = 0 has one (coincidental) root when "b^2 = 4ac"
16 = 4k^2
k = +2 or -2 so those equations are:-
2x^2 - 4x + 2 = 2(x - 1)^2 = 0 with one root at x = 1
-2x^2 - 4x - 2 = -2(x + 1)^2 = 0 with one root at x = -1

Regards - Ian