Question

Akar akar persamaan kuadrat x^2+(p-3)x+4=0 adalah x1 dan x2. jika x1^2 + x2^2 = p-5, nilai p yang memenuhi adalah

Answer 1

(x1+x2)² = x1² + 2x1x2 + x2²

Itu artinya, x1² + x2² = (x1+x2)² - 2x1x2

x² + (p-3)x + 4 = 0
a = 1
b = p-3
c = 4

x1 + x2 = -b/a
= -(p-3) / 1
= -p + 3

x1x2 = c/a
= 4/1
= 4

(x1+x2)² - 2x1x2 = p - 5
(-p + 3)² - 2.4 = p - 5
p² - 6p + 9 - 8 - p + 5 = 0
p² - 7p + 6 = 0

(p - 6)(p - 1)

p - 6 = 0
p = 6

p - 1 = 0
p = 1

Jadi p = 6 atau p = 1

Answer 2

Jadi, p = 6 atau p = 1