himpunan jawab pertidaksamaan |x - 2|² - 4|x - 2| < 12

|x - 2|² - 4|x - 2| < 12
= |x - 2|² - 4|x - 2|-12 < 0

misal y=|x - 2|

y²-4y-12<0
(y-6)(y+2)=0

#y+2=0
y=-2 => |x-2|=-6 (bkan penyelsaian mutlak selalu postif)

#y-6=0
y=6 => |x-2|=6

~ x-2=6
x=8

~x-2=-6
x=-4

Hp={x| -4<x<8}

|x - 2|² - 4|x - 2| < 12
|x - 2|² - 4|x - 2| - 12 < 0
Misal |x - 2| = a
a² - 4a - 12 < 0
(a - 6)(a + 2) < 0
a = 6 atau a = -2
Garis bilangan
+++ (-2) --- (6) +++
-2 < a < 6
-2 < |x - 2| < 6
Karena |x - 2| selalu bernilai positif maka
0 < |x - 2| < 6
|x - 2| < 6
(x - 2)² < 6²
x² - 4x + 4 < 36
x² - 4x - 32 < 0
(x - 8)(x + 4) < 0
x = 8 atau x = -4
Garis bilangan
+++ (-4) --- (8) +++
-4 < x < 8