Question on: WAEC Mathematics - 2013
Find the values of k in the equation 6k2 = 5k + 6
A
{\(\frac{-2}{3}, \frac{-3}{2}\)}
B
{\(\frac{-2}{3}, \frac{3}{2}\)}
C
{\(\frac{2}{3}, \frac{-3}{2}\)}
D
{\(\frac{2}{3}, \frac{3}{2}\)}
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Correct Option: B
6k2 = 5k + 6
6k2 - 5k - 6 = 0
6k2 - 0k + 4k - 6 = 0
3k(2k - 3) + 2(2k - 3) = 0
(3k + 2)(2k - 3) = 0
3k + 2 = 0 or 2k - 3 = 0
3k = -2 or 2k = 3
k = \(\frac{-2}{3}\) or k = \(\frac{3}{2}\)
k = (\(\frac{-2}{3}\), k = \(\frac{3}{2}\))
6k2 - 5k - 6 = 0
6k2 - 0k + 4k - 6 = 0
3k(2k - 3) + 2(2k - 3) = 0
(3k + 2)(2k - 3) = 0
3k + 2 = 0 or 2k - 3 = 0
3k = -2 or 2k = 3
k = \(\frac{-2}{3}\) or k = \(\frac{3}{2}\)
k = (\(\frac{-2}{3}\), k = \(\frac{3}{2}\))
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