diketahui akar-akar persamaan kuadrat x^2 - px + p + 1 =0 adalah x1 dan x2.jika x1^2 + x2^2=6 maka Nilai p

x^2 - px + p + 1 = 0
a = 1,b= -p ,c = p+1

x1+x2 = -b/a = p/1 = p
x1.x2 = c/a = p+1/1 = p+ 1

x1^2 + x2^2 = 6
(x1+x2)^2 - 2x1.x2 = 6
(p)^2 - 2(p+1) = 6
p^2 - 2p -2 - 6 = 0
p^2 -2p -8 = 0
(p-4)(p+2)
p = 4 dan p=-2

x² - px + p + 1 = 0, dengan a = 1, b = -p dan c = p + 1

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

Maka:
x1² + x2² = 6
6 = (x1 + x2)² - 2(x1.x2)
6 = p² - 2(p + 1)
6 = p² - 2p - 2
p² - 2p - 2 - 6 = 0
p² - 2p - 8 = 0
(p - 4)(p + 2) = 0
p = 4 dan p = -2

Jadi, nilai p adalah -2 atau 4