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diatas sb :
D < 0
b²-4ac < 0
(k-4)² - 4 k 1/2 < 0
k² - 8k + 16 - 2k < 0
k² - 10k + 16 < 0
(k - 8) (k - 2) = 0
k=8 atau k=2
D < 0 "Rumus D < 0 = b²-4ac < 0
b²-4ac < 0
(k-4)² - 4 k 1/2 < 0
k² - 8k + 16 - 2k < 0
k² - 10k + 16 < 0
(k - 8) (k - 2) = 0
k=8 atau k=2