If 6k2 + k = 2 and k > 0, then k must equal which of the following?
6k2 + k = 2
6k2 + k − 2 = 0
This will factor into some (ak + b)(ck − d) such that ac = 6, bd = 2, and bc − ad = 1. The combination that works is
(3k + 2)(2k − 1) = 0
3k + 2 = 0 OR 2k − 1 = 0
k = OR k =
Since we know k > 0, it must be that k = .
Answer = (A)
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