To solve absolute value inequalities, the following properties can be used:
Form 1
(a). Jika │f(x)│ < a maka –a < f(x) < a
(b). If │f (x) │> a then f (x) <–a or f (x)> a
Form 2
(a). If │f (x) │ <g (x) then f 2 (x) <g 2 (x). The condition is g (x)> 0
(b). If │f (x) │> g (x) then f 2 (x)> g 2 (x). The condition is g (x)> 0
Form 3
(a). Jika │f(x)│ < │g(x)│ maka f2(x) < g2(x).
(b). Jika │f(x)│ > │g(x)│ maka f2(x) > g2(x).
To better understand the absolute value inequality, consider the following example:
01. Determine the interval of the value of x that satisfies │2x + 3│ <5
Answer
│2x + 3│ <5
–5 < 2x + 3 < 5
–5 – 3 < 2x + 3 – 3 < 5 – 3
–8 < 2x < 2
–4 < x < 1
02. Determine the interval of the value of x that satisfies │2 – 3x│ <8
Answer
│2 – 3x│ <8
–8 < 2 – 3x < 8
–8 – 2 < 2 – 3x – 2 < 8 – 2
–10 < –3x < 6
10/3 > x > –2
–2 < x < 10/3
03. Determine the interval of the value of x that satisfies │2x + 6│> 4
Answer
│2x + 6│ > 4
2x + 6 <–4 or 2x + 6> 4
2x <–4 – 6 or 2x> 4 – 6
2x <–10 or 2x> –2
x <–5 or x> –1
04. Determine the interval of the value of x that satisfies │5 – 3x│> 4
Answer
│5 – 3x│ > 4
5 – 3x <–4 or 5 – 3x> 4
–3x <–4 – 5 or –3x> 4 – 5
–3x <–10 or –3x> –1
x> 10/3 or x <1/3
x <1/3 or x> 10/3
05. Determine the interval of the value of x that satisfies │3x – 2│ <2x + 7
Answer
│3x – 2│ < 2x + 7
(3x – 2)2 < (2x + 7)2
9x2 – 12x + 4 < 4x2 + 28x + 49
9x2 – 12x + 4 – 4x2 – 28x – 49 < 0
5x2 – 40x – 45 < 0
x2 – 8x – 9 < 0
(x – 9)(x + 1) < 0
x = 9 and x = –1 so that: –1 <x <9 ……………… (1)
Conditions: 2x + 7> 0
2x > –7
x > –7/2 ………………………………………………………… (2)
from (1) and (2) the interval is obtained: –1 <x <9
06. Determine the interval of the value of x that satisfies │2x – 9│ <4x – 3
Answer
│2x – 9│ <4x – 3
(2x – 9) 2 <(4x – 3) 2
4x2 – 36x + 81 < 16x2 – 24x + 9
4x2 – 36x + 81 – 16x2 + 24x – 9 < 0
–12x2 – 12x + 72 < 0
x2 + x – 6 > 0
(x – 2)(x + 3) > 0
x = 2 and x = –3 so that: x <–3 or x> 2 …………… (1)
Conditions: 4x – 3> 0
4x > 3
x > 3/4 …………………………………………………………………. (2)
from (1) and (2) the interval: x> 2 is obtained
07. Determine the interval of the value of x that satisfies │x + 4│ ≥ │3x – 8│
Answer
X + 4│ │3x – 8│
(x + 4)2 ≥ (3x – 8)2
x2 + 8x + 16 ≥ 9x2 – 48x + 64
x2 + 8x + 16 – 9x2 + 48x – 64 ≥ 0
–8x2 + 56x – 48 ≥ 0
x2 – 7x + 6 ≤ 0
(x – 6)(x – 1) ≤ 0
x = 6 and x = 1 so that: 1 ≤ x ≤ 6