Question

Akar-akar persamaan x^2+4x-1=0 adalah x1danx2 persamaan kuadrat yang akar-akarnya (2x1-3) dan (2x2-3) adalah....

Answer 1

X^2 + 4x - 1 = 0
x1 + x2 = -b/a = -4/1 = -4
x1 . x2 = c/a = -1/1 = -1

misal 2x1 - 3 = p dan 2x2 - 3 = q

p + q = (2x1 - 3) + (2x2 - 3)
= 2x1 + 2x2 - 3 - 3
= 2(x1 + x2) - 6
= 2(-4) - 6
= -14

pq = (2x1 - 3)(2x2 - 3)
= 4x1.x2 - 6x1 - 6x2 + 9
= 4x1.x2 - 6(x1 + x2) + 9
= 4(-1) - 6(-4) + 9
= 29

pers kuadrat baru
x^2 - (p+q)x + pq = 0
x^2 - (-14)x + 29 = 0
x^2 + 14x + 29 = 0

CARA LAIN
2x - 3 = a => 2x = a + 3 => x = (a + 3)/2
x^2 + 4x - 1 = 0
=> [(a + 3)/2]^2 + 4(a + 3)/2 - 1 = 0
=> (a^2 + 6a + 9)/4 + 2a + 6 - 1 = 0
=> a^2 + 6a + 9 + 8a + 20 = 0
=> a^2 + 14a + 29 = 0
=> x^2 + 14x + 29 = 0