Question

Diketahui akar akar persamaan kuadrat X² + (P-1)X – 5 =0 adalah a dan b ,jika a²+b²-2ab =8p , maka tntukan nilai p ??????tolong jwab beserta uraian.a

Answer 1

Ingat bahwa (a - b)² = a² + b² - 2ab

x² + (p - 1)x - 5 = 0

a = 1
b = p - 1
c = -5

a - b = √D / a
= √(b² - 4ac) / a
= √((p - 1)² - 4(1)(-5)) / 1
= √p² - 2p + 1 + 20
= √p² - 2p + 21

(√p² - 2p + 21)² = 8p
p² - 2p + 21 = 8p
p² - 2p - 8p + 21 = 0
p² - 10p + 21 = 0
(p - 7)(p - 3) = 0

p = 7 atau p = 3

Answer 2

a²+b² = (a+b) -2ab


a² + b² - 2ab = 8p
(a+ b)² - 2ab - 2ab = 8p
(a +b )² - 4ab = 8p
[-(p-1)/1]² - 4( -5/1) = 8p
(p-1)² +20 = 8p
p² -2p +1 +20 =8p
p² -10p +21 = 0
(p -7)(p-3) = 0
p =7 atau p=3