Akar akar dari persamaan x³-14x²+ax+b=0membentuk deket geometri dengan rasio 2.tentukan nilai a dan b.

jawab

x³ - 14x² + ax + b = 0 , akar akarnya x1,x2, x3
x1 + x2 + x3 = - (-14)/(1) = 14
x1 x2 + x1 x3 + x2 x3 = a/1 = a
x1 x2 x3 = - (b/1) = - b

x1,x2,x3 membentuk deret geo dengan r= 2, maka
x1 = x1
x2 = 2x1
x3 = 4x1

x1 + x2 + x3 = 14
x1+ 2x1 + 4x1 = 14
7x1 = 14
x1 = 2

x1 x2 + x1 x3 + x2 x3 = a
x1 (2x1) + x1 (4x1) + (2x1)(4x1) = a
2x1² + 4x1² + 8x1² = a
14 x1²= a
a = 14 (2)²
a = 56

x1 x2 x3 = - b
x1 (2x1)(4x1) = - b
8 x1³ = - b
- b = 8(2)³
- b = 64
b = - 64


nilai a = 56
nilai b =  - 64