Question

Persamaan Kuadrat x2-6x+3=0 mempunyai akar-akar p dan q. Tentukan PK baru yg akar-akarnya 3p+1 dan 3q+1

Answer 1

a = 1: b = -6 ; c = 3
p+q = -b/a = 6
p×q = c/a = 3
y1 = 3p+1
y2 = 3q + 1

x² - (3p+1 + 3q+1)x + (3p+1)(3q+1) = 0
x² - {3p + 3q + 2)x + 9pq + 3p + 3q + 1 = 0
x² - { 3(p+q) + 2}x + 9 (p×q) + 3(p+q) + 1 = 0
x² - (3(6) + 2)x + 9 (3) + 3(6) + 1= 0
x² - 20x + 46 = 0
semoga membantu :)