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a = 1, b = -(m+3), c= 3m +1
1 akar riil maka D = 0
D = 0
b² - 4ac = 0
(- (m+3))² - 4(1) (3m+1) = 0
m² + 6m + 9 - 12m - 4 = 0
m² - 6m + 5 = 0
(m - 5) (m - 1) = 0
m =5 atau m = 1
x²-(m+3)x + (3m+1) = 0
a = 1 , b = -(m+3) , dan c = 3m + 1
D = 0
b² - 4a c = 0
(-(m+3))² - 4 . 1 (3m+1) = 0
m²+6n+9-12m-4 = 0
m² -6m + 5 = 0
(m-1)(m-6) = 0
jadi batasnya m = 1 dan m = 6
jadi nilai m yang memenuhi agar mempunyai akar persamaan riil yaitu m=1 atau m=6