Question

Diketahui persamaan pangkat tiga dengan 2x³-px²+7x-2=0, dua diantara akar-akarnya saling berkebalikan. Hitunglah nilai p yang memenuhi persamaan tersebut!

Answer 1

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Jawb

2x³- px² + 7x  - 2 = 0
a = 2
b = -p
c= 7
d = -2
akar akarnya x1,x2 atau x3
x1 + x2 + x3 = -b/a = p/2
x1. x2 + x1. x3 + x2. x3 = c/a = 7/2
x1. x2. x3 = -d/a = 1

dua akar saling berkebalikan , misal x1 = 1/x2 --> maka x1. x2 = 1
x1. x2. x3 = 1
(1). x3= 1
x3 = 1

x1 + x2 + x3 = p/2
x1 + x2 = p/2 - 1 ....(i)

x1 x2 + x1. x3  + x2. x3 = 7/2
(1)  + x1 + x2 = 7/2
x1 + x2 = 7/2 - 1 = 5/2 ...(ii)

(i dan ii)
x1 + x2 = p/2 - 1
x1 + x2 = 5/2 

p/2 - 1 = 5/2
p -  2 = 5
p = 7