Question

Akar2 persamaan ax4-10x3+bx-3=0 adalah x1,x2,x3 dan x4 dengan x1<x2<x3<x4. Jika dua akarnya adalah-1 dan 3, maka 6x2-x3-2x1+x4= ...

Answer 1

Ax⁴ - 10x³ + bx - 3= 0

Akar --> -1
a(-1)⁴ - 10(-1)³ + b(-1) - 3 = 0
a + 10 - b - 3 = 0
a - b = - 7 ... (1)

Akar --> 3
a(3)⁴ - 10(3)³ + b(3) - 3 = 0
81a - 270 + 3b - 3 = 0
81a + 3b = 273 (: 3)
27a + b = 91 (x -1)
-27a - b = -91 ... (2)

Eliminasi pers (1) dan (2) :
a - b = - 7
-27a - b = -91 -
28a = 84
a = 3
b = 10

Persamaannya 3x⁴ - 10x³ + 10x - 3 = 0
Gunakan Metode Horner mencari faktor-faktornya :
x3 = 1/3
x4 = 1
Bukti :
Masukkan x4 = 1
3(1)⁴ - 10(1)³ + 10(1) - 3 = 0
3 - 10 + 10 - 3 = 0
0 = 0 (TERBUKTI)

Karena x1<x2<x3<x4 maka :
x1 = -1
x2 = 1/3
x3 = 1
x4 = 3

Maka 6x2 - x3 - 2x1 + x4 = 6(1/3) - 1 - 2(-1) + 3 = 2 - 1 + 2 + 3 = 6

CMIIW :)