Posted by Neha on Sunday, October 21, 2012 at 7:23pm.
determine the value of k for which the fuction f(x)=4x^23x+2kx+1 has 2 zeros.

Math  Reiny, Sunday, October 21, 2012 at 9:24pm
To have 2 zeros, or two real solutions, the discriminant must be positive
for 4x^2  3x + 2kx + 1
a = 4
b = 2k3
c = 1
b^2  4ac > 0
(2k3)^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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