diketahui (x-3), (x-1), (2x-2), dan (2x+6) membentuk barisan geometri. Nilai x+2=

dik : (x - 3) (x-1) (2x-2) (2x+6)
dit : x +2 =

U2/U1 = U3/U2 
U2^2 = (U1) (U3)
(x-1)^2 = (x-3) (2x-2)
(x-1)(x-1) = (x-3) (2x-2)
x^2 - x - x +1 = 2x^2 - 2x - 6x + 6
x^2 - 2x^2 +1 = 2x^2 - 8x + 6
(x^2 - 2x^2) - 2x + 8x +1 -6 = 0
-x^2 + 6x - 5 = 0
(x^2 - 6x + 5) = 0
(x - 1) (x - 5) = 0

x= 1 atau x = 5
misalnya x = 1 ----- x +2 = 3
               x = 5 ------x +2 = 7
jawabannya ada 2