jika xkuadrat+2mx+(5m-4)=0 memiliki akar kembar.nilai m adalah

(-6)^2 - 4(3)(2p) = 132
36 - 24p = 132
96 = 24p
p = 4

Akar kembar ---> D = 0
(2m)^2 - 4(1)(5m - 4) = 0
4m^2 - 20m + 16 = 0
------------------------------ (bagi 4)
m^2 - 5m + 4 = 0
(m - 1)(m - 4) = 0
m = 1 ; m = 4