Question

Tentukan persamaan kuadrat yang akar akarnya dua lebihnya dari akar akar persamaan kuadrat 2x^2 + 5x - 1 =0.

Answer 1

2x² + 5x - 1 = 0
maka :
a = 2
b = 5
c = -1
jadi
x1 + x2 = -b/a = -5/2
x1.x2 = c/a = -½

akar - akar baru :
(x1 + 2) dan (x2 + 2)
maka :
x1+x2 = (x1+2)+(x2+2)
= (x1+x2) + 4
= -5/2 + 4
= -5/2 + 8/2
= 3/2
x1.x2 = (x1+2)(x2+2)
=x1.x2 + 2x1+2x2 + 4
=x1.x2 + 2(x1+x2) + 4
= -½ + 2.3/2 + 4
= -½ + 3 + 4
= - ½ + 7
= -½ + 14/2
= 13/2

PK baru :
x² - (x1+x2)x + x1.x2 = 0
x² - 3/2 x + 13/2 = 0 --> ×2
2x² - 3x + 13 = 0