Akar-akar persamaan kuadrat X^ +PX+6=0, adalah X1 & X2 .Jika X1 & X2 positif,serta memenuhi X1=3/2 X2,, maka nilai p =?? Please,tolong dijawab

jawab

x² +px + 6
(i) x1 + x2= -p
(ii) x1 x 2= 6
(ii) x1 = 3/2 x2


(ii) dan (ii)
x1 x2 = 6
x1 = 3/2 x2

3/2 x2. x2 = 6
x2² = 6 : 3/2 = 4
x2= 2 atau x2 = -2

x2= 2 --> x1 = 3
x2= -2 --> x2 = - 3
x1 dan x2  positif
x1= 2 , x1 = 3
x1 + x2 = - p
p = -(x1+x 2)
p= - (2 + 3)
p = - 5