Selesalkan Persamaan Kundrat Jergan Faktorvosi 2 1. x² -7 x + 6 = 0 2 b. x² + 3x -4 = 0 2 C. X² - 2x - 48 = 0.
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Jawab:
(1.) Hp = {6, 1}
(2.) Hp = {-4, 1}
(3.) Hp = {8, -6}
Penjelasan:
Rumus:
Ax²+Bx+C = 0
Jika p dan q akar-akarnya
pq = C/A, p+q = -B/A
(1.) x²-7x+6 = 0
A = 1, B = -7, C = 6
Teori Vieta, p dan q Hp-nya
pq = C/A, p+q = -B/A
pq = 6/1, p+q = -(-7)/1
pq = 6, p+q = 7
p = 6, q = 1
Hp = {6, 1}
(2.) x²+3x-4 = 0
A = 1, B = 3, C = -4
Teori Vieta, p dan q Hp-nya
pq = C/A, p+q = -B/A
pq = -4/1, p+q = -3/1
pq = -4, p + q = -3
p = -4, q = 1
Hp = {-4, 1}
(3.) x²-2x-48 = 0
A = 1, B = -2, C = -48
Teori Vieta, p dan q Hp-nya
pq = C/A, p+q = -B/A
pq = -48/1, p+q = -(-2/1)
pq = -48, p + q = 2
p = 8, q = -6
Hp = {8, -6}
(xcvi)
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