Jika akar akar dari 3^x2 + ax- 2 =0 dan 2^x2 + 6x + 3b=0 saling berkebalikan, Maka hasil dari b-a adalah?

saling berkebalikan x1 jadi 1/x1 dan x2 jadi 1/x2
3x² + ax - 2 = 0
akarnya x1 dan x2
x1 + x2 = -b/a = -a/3
x1.x2 = c/a = -2/3

2x² + 6x + 3b = 0
akarnya 1/x1 dan 1/x2
1/x1 + 1/x2 = -b/a = -6/2 = -3
(x1 + x2)/(x1.x2) = -3
(-a/3)/(-2/3) = -3
(-a/3) × (3/-2) = -3
a/2 = -3
a = -6
1/x1 . 1/x2 = 3b/2
1 / (x1.x2) = 3b / 2
1 / (-2/3) = 3b / 2
3b × (-2/3) = 1 × 2
-2b = 2
b = -1

b - a = -1 - (-6) = -1 + 6 = 5