Question

Persamaan 3x^3 - 4x^2 + bx + 6 = 0 mempunyai sepasang akar akar yang berkebalikan. tentukan nilai b dan akar akar persamaan tersebut

Answer 1

X1 = 1/x2
x1.x2 = 1

x1.x2.x3 = -D/A
1.x3 = -6/3
x3 = -2

x1 + x2 + x3 = -B/A
x1 + x2 + (-2) = 4/3
x1 + x2 = 1 1/3 + 2
x1 + x2 = 3 1/3 = 10/3

x1.x2 + x1.x3 + x2.x3 = C/A
1 + (-2x1) + (-2x2) = b/3
1 - 2x1 - 2x2 = b/3
1 -2 (x1+x2) = b/3
1 -2(10/3) = b/3
1 - 20/3 = b/3. -> kedua ruas di kali 3
3 - 20 = b
-17 = b
jadi, b = -17