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December 19, 2014

December 19, 2014

Posted by **Annie** on Wednesday, January 25, 2012 at 7:44pm.

- math -
**Reiny**, Wednesday, January 25, 2012 at 10:17pmto have two zeros, either real or complex,

b^2 - 4ac ≠ 0

4x^2 + x(2k - 3) + 1

discrim. = (2k-3)^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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