Posted by Annie on Wednesday, January 25, 2012 at 7:44pm.
Determine the values of k for which the function f(x)= 4x^23x+2kx+1 has two zeros , check these values in the original equation

math  Reiny, Wednesday, January 25, 2012 at 10:17pm
to have two zeros, either real or complex,
b^2  4ac ≠ 0
4x^2 + x(2k  3) + 1
discrim. = (2k3)^2  4(4)(1)
= 4k^2  12k + 9  16
= 4k^2  12k  7
to have only 1 root
4k^2  12k  7 = 0
(2k  7)(2k + 1) = 0
k = 7/2 or k = 1
So to have two roots, k ≠ 7/2 or k ≠ 1
So put any value of k other than those two and you will get 2 differentroots.
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