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July 31, 2014

July 31, 2014

Posted by **Neha** on Sunday, October 21, 2012 at 7:23pm.

- Math -
**Reiny**, Sunday, October 21, 2012 at 9:24pmTo 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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