contoh Salah soal satu akar persamaan suku banyak

x^2- (3+2a) x^2+(a^2+5a=2)x-2a(a+1)=0 adalah 2. Selisih antara akar-akar?
itu salah mungkin,x^2 diganti x^3 , trus = diganti +
masukkan x = 2
2^3 - (3 + 2a) 2^2 + (a^2 + 5a + 2) 2 - 2a(a+1) = 0
8 - (12 + 8a) + (2a^2 + 10a + 4) - 2a^2 - 2a = 0
8 - 12 - 8a + 2a^2 + 10a + 4 - 2a^2 - 2a = 0
habis semua, maka a = 0
suku banyak baru :
x^3 - 3x^2 + 2x = 0

x1 + x2 + x3 = -b/a
2 + x2 + x3 = 3/1
x2 + x3 = 1
x2 = 1 - x3

(x1x2) + (x1x3) + (x2x3) = c/a
2(1 - x3) + 2x3 + (1 - x3) x3 = 2/1
2 - 2x3 + 2x3 + x3 - x3^2 = 2
x3^2 - x3 = 0
x3 (x3 - 1) = 0
x3 = 0 v x3 = 1

x2 = 0 v x2 =1

maka selisihnya :
x3 - x2 = 1 - 0 = 1