Question

Jika x1 dan x2 adalah akar - akar PK x2 - 5x + 6 =0 tentukan PK baru yang akar-akarnya :
a) berkebalikan dengan akar PK lama
b) berlawanan dengan akar PK lama
c) 2 kurangnya dari
d) 1/x1 dan -1/x2

Answer 1

Komposisi:
x₁+x₂ = 5 (Dari -b/a)
x₁x₂ = 6 (Dari c/a)

a.)
(x-1/x₁)(x-1/x₂) = 0
x² - (1/x₁ + 1/x₂)x + 1/(x₁x₂) = 0
x² - [(x₁+x₂)/(x₁x₂)]x + 1/(x₁x₂) = 0
(x₁x₂)x² - (x₁+x₂)x + 1 = 0
Substitusikan:
6x² - 5x + 1 = 0

b.)
(x-(-x₁))(x-(-x₂)) = 0
(x+x₁)(x+x₂) = 0
x²+(x₁+x₂)x+x₁x₂ = 0
Substitusikan:
x² + 5x + 6 = 0

c.)
2 kurangnya dari:
(x-(x₁-2))(x-(x₂-2)) = 0
x²-(x₁-2+x₂-2)x + (x₁-2)(x₂-2) = 0
x² - (x₁+x₂-4)x + x₁x₂-2(x₁+x₂)+4 = 0
Substitusikan:
x² - (5-4)x + 6-2(5)+4 = 0
x² - x = 0

d.)
Asumsi:
-1/x₁ dan 1/x₂
Maka,
(x-(-1/x₁))(x-(-1/x₂)) = 0
(x+1/x₁)(x+1/x₂) = 0
x² + (1/x₁+1/x₂) + 1/(x₁x₂) = 0
x² + [(x₁+x₂)/(x₁x₂)]x + 1/x₁x₂ = 0
(x₁x₂)x² + (x₁+x₂)x + 1 = 0
Substitusikan:
6x² + 5x + 1 = 0

Answer 2

Identifikasi akar jumlah Dan akar Kali x1+x2= -b/a = -(-5)/1 = 5 x1.x2 = c/a = 6/1 = 6 rumus PK baru : x²-(A+B)x+(A.B)= 0 a. berkebalikan A = 1/x1 B = 1/x2 A+B = 1/x1 + 1/x2 = x1+x2 / x1.x2 = 5/6 A.B = 1/x1. 1/x2 = 1/6 PK : x²- (5/6) x + 1/6 = 0 .... x6 6x²- 5x + 1 = 0 b. berlawanan A = -x1 B = -x2 A+B = -x1+(-x2) = - (x1+x2) = - 5 A.B = -x1.-x2 = 6 PK : x² - (-5)x + 6 = 0 x² + 5x + 6 = 0 c., 2 kurangnya A = x1-2 B = x2 - 2 A+B = x1+x2-4 = 5-4 = 1 A.B = (x1-2)(x2-2) = x1.x2 - 2(x1+x2) + 4 = 6 - 2(5)+4 = 0 PK : x² - x + 0 = 0 x² - x = 0 d. A = 1/x1 B = - 1/x2 A+B = 1/x1 - 1/x2 = x2-x1 / x1.x2 = - (x1-x2) / x1.x2 jika (x1-x2)² = x1²-2x1.x2+x2² = (x1+x2)²-4x1.x2 x1-x2 = √ (x1+x2)²-4x1.x2 = √ 5²-4.6 = 1 sehingga A+B = - 1/6 A.B = 1/x1. (- 1/x2) = - 1/ x1.x2 = - 1/6 PK : x² - (-1/6)x + (- 1/6) = 0....x6 6x² + x -1 = 0