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x² + (2k+4)x + 9k = 0
b²-4ac=0
(2k+4)² - 4(1)(9k)=0
4k²+16k+16 -36k=0
4k² - 20k +16 =0
k² -5k +4 =0
(k-4)(k-1)=0
k=4 atau k=1