Posted by Neha on Sunday, October 21, 2012 at 7:23pm.
To have 2 zeros, or two real solutions, the discriminant must be positive
for 4x^2 - 3x + 2kx + 1
a = 4
b = 2k-3
c = 1
b^2 - 4ac > 0
(2k-3)^2 - 4(4)(1) > 0
4k^2 - 12k -7 > 0
(2k + 1)(2k - 7) > 0
k < -1/2 OR k > 7/2
You might want to put that into the interval notation you have been taught.
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