salah satu akar persamaan kuadrat ax^2+(a+1)x+(a-1)=a>0adalahx1. jika akarlainnya x2=2x1 maka konstanta a?

ax^2 + (a+1)x + (a-1) = 0
A = a, B = a+1, C = a-1
x1+x2 = -B/A
x1 + 2x1 = -(a+1)/a
3x1 = -(a+1)/a
x1 = -(a+1)/3a

x1.x2 = C/A
x1. 2x1 = (a-1)/a
x1 = (a-1)/2a

x1 = x2
-(a+1)/3a = (a-1)/2a
-(a+1)/3 = (a-1)/2
-2(a+1) = 3(a-1)
-2a -2 = 3a-3
1 = 5a
a = 1/5