To understand the concept of absolute value, the following story will be illustrated: A boy scout is practicing marching. From a stationary position, the child is asked to take 2 steps forward, then 4 steps back. Followed by 3 steps forward and finally 2 steps back. From the story above, the following problems can be drawn:
a. How many steps did the boy scout take from the first to the last?
b. Where was the scout’s last position, if measured from a stationary position? (how many steps forward or how many steps back)
To answer the above problem, the following number line will be given:
From the picture above, we assume that x = 0 is the rest (initial) position of the child. The arrow to the right indicates the direction of the step forward (a positive value) and the arrow to the left indicates the direction of the step backward (a negative value). So that the above problems can be answered as follows:
a. The number of steps of the boy scout from the first to the last is the addition form 2 + 4 + 3 + 2 = 11 steps. This form of addition is an addition without paying attention to the forward (positive) and backward (negative) directions.
b. From the picture above, it can be seen that the last position of the boy scout, when measured from a stationary position is 1 step back (x = –1). This result is obtained from the addition of 2 + (–4) + 3 + (–1) = –1. This form of addition is an addition by paying attention to the forward (positive) and backward (negative) directions.
The illustration of solving problem (a) above is the basis of the concept of absolute value. Where the absolute value of a real number x is the distance between that number and zero on the number line. And denoted by x│. Formally absolute value is defined
From the above concept is obtained: │ – 3│ = 3, │ – 15│ = 15, │6│ = 6, │10│ = 10 and so on.
To better understand absolute value inequalities, consider the following example:
01. Determine the value of
(a) │–4│ + │5│ – │–3│ (b) 6 – │–2│ + │–5│ + 1
(c) │4 – │–7││ (d) │–9 + │–2││
Answer
(a) │–4│ + │5│ – │–3│ = 4 + 5 – 3 = 6
(b) 6 – │–2│ + │–5│ + 1 = 6 – 2 + 5 + 1 = 10
(c) │4 – │–7││ = │4 – 7│ = │–3│ = 3
(d) │–9 + │–2││ = │–9 + 2│ =│–7│ = 7
02. For x = –3, then determine the value of x2 + 6x + 5│
Answer
│x2 + 6x + 5│ = │(–3)2 + 6(–3) + 5│
= │9 – 18 + 5│
= │–4│
= 4
03. For x = 2, then determine the value 4│2 – 6x│+ 3x – 8│
Answer
4│2 – 6x│+ │3x – 8│ = 4│2 – 6(2)│+ │3(2) – 8│
= 4│–10│+ │–2│
= 40 + 2
= 42
04. For x = –2, then determine the value of x2 – 6x│– 4x + 5│
Answer
│x2 – 6x│– │4x + 5│= │(–2)2 – 6(–2)│– │4(–2) + 5│
= │4 + 12│– │–8 + 5│
= 16 + 3
= 19
05. A snail will climb the flagpole starting early August 5th. If on an odd date the snail rises 5 m high, and on an even date it descends 3 m, then it will reach the top of the flagpole at the end of August 17th.
(a) What is the height of the flagpole
(b) How far did the snail travel?
Answer
(a) Height of flagpole = 5 – 3 + 5 – 3 + 5 – 3 + 5 – 3 + 5 – 3 + 5 – 3 + 5 = 17m
(b) the distance traveled by the snail = 5 + –3│ + 5 + –3│+ 5 + –3│+ 5 + –3│+ 5 + –3│ + 5 + –3│+ 5 = 53 m
To solve the absolute value equation, properties can be used
01. (a) Jika │f(x)│ = a maka f2(x) = a2
(b) If f(x)│ = a then f(x) = a or f(x) = –a
02. (a) Jika │f(x)│ = │g(x)│ maka f2(x) = g2(x)
(b) If f(x)│ = g(x)│ then f(x) = g(x) or f(x) = –g(x)
To better understand absolute value inequalities, consider the following example:
01. Determine the value of x that satisfies the following equation:
(a) │2x – 5│ = 3 (b) │4 – 3x│ = 6
Answer
(a) │2x – 5│ = 3
(2x – 5) 2 = 32
4x2 – 20x + 25 = 9
4x2 – 20x + 16 = 0
x2 – 5x + 4 = 9
(x – 4)(x – 1) = 0
So x = 1 and x = 4
(b) │3 – 2x│ = 7
(3 – 2x) 2 = 72
9 – 12x + 4x2 = 49
4x2 – 12x – 40 = 0
x2 – 3x – 10 = 0
(x – 5)(x + 2) = 0
So x = 5 or x = –2
02. Determine the value of x that satisfies the following equation:
(a) │2x + 4│ = │x – 1│ (b) │3x + 4│ = │2x – 1│
Answer
(a) │2x + 4│ = │x – 1│
(2x + 4) 2 = (x – 1) 2
4x2 –16x + 16 = x2 – 2x + 1
3x2 – 14x + 15 = 0
(3x – 5)(x – 3) = 0
So x = 5/3 or x = 3
(b) │3x + 4│ = │2x – 1│
(3x + 4) 2 = (2x – 1) 2
9x 2 + 24x + 16 = 4x 2 – 4x + 1
5x2 + 28x + 15 = 0
(5x + 3)(x + 5) = 0
So x = –3/5 or x = –5
03. Determine the value of x that satisfies the following equation:
(a) │3x – 2│ = x + 4 (b) │2x + 4│ = x – 3
Answer
(a) │3x – 2│ = x + 4
(3x – 2)2 = (x + 4)2
9x2 – 12x + 4 = x2 + 8x + 16
8x2 – 20x – 12 = 0
2x2 – 5x – 3 = 0
(2x + 1) (x – 3) = 0
So x = –1/2 or x = 3
Test: x = –1/2 then x + 4 = –1/2 + 4 = 7/2 (meet)
Test: x = 3 then x – 4 = 3 + 4 = 7 (meet)
Until H = {–1/2, 3}
(b) │2x – 4│ = x – 3
(2x – 4) 2 = (x – 3) 2
4x2 –16x + 16 = x2 – 6x + 9
3x 2 – 22x + 7 = 0
(3x – 1)(x – 7) = 0
So x = 1/3 or x = 7
Test: x = 1/3 then x – 3 = 1/3 – 4 = –11/3 (does not meet)
Test: x = 7 then x – 4 = 7 – 4 = 3 (meet)
Until H = {7}