" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
a=1 ; b=-(2+2m) ; c=3m+3
Syarat akar tdk real : D<0
D = Diskriminan
D = b^2 - 4ac
D < 0
(-(2+2m))^2 - 4(1)(3m+3) < 0
(4×(1+m)^2) - 4×3×(1+m) < 0
4×(1+m)[(1+m) - 3] < 0
4(1+m)(m-2) < 0
m=-1 U m=2
+++++ ---------- ++++++
-1 2
Tanda di pertidaksamaan di 4(1+m)(m-2) < 0 adalah lebih kecil dari 0.
daerah yg memenuhi daerah negatif : -1
Semoga membantu