MA 3972-MA-Book May 8, 2018 13:4650 STEP 4. Review the Knowledge You Need to Score High
5.2 Absolute Values and Inequalities
Main Concepts:Absolute Values, Inequalities and the Real Number Line, Solving Absolute
Value Inequalities, Solving Polynomial Inequalities, and Solving Rational
InequalitiesAbsolute Values
Letaandbbe real numbers.- |a|=
{
a,ifa≥ 0
−a,ifa< 0- |ab|=|a||b|
- |a−b|=|b−a|
4.√
a^2 =|a|={
a,ifa≥ 0
−a,ifa< 0Example 1
Solve forx:| 3 x− 12 |=18.
Depending on whether the value of (3x−12) is positive or negative, the equation
| 3 x− 12 |=18 could be written as 3x− 12 =18 or 3x− 12 =−18. Solving both equations,
you havex=10 andx=−2. (Be sure to check both answers in the original equation.) The
solution set forxis{−2, 10}.Example 2
Solve forx:| 2 x− 12 |=| 4 x+ 24 |.
The given equation implies that either 2x− 12 = 4 x+24 or 2x− 12 =−(4x+24). Solv-
ing both equations, you havex=−18 andx=−2. Checkingx=−18 with the original
equation:|2(−18)− 12 |=|4(−18)+ 24 |or|− 36 − 12 |=|− 72 + 24 |or|− 48 |=|− 48 |.
Checkingx=−2 with the original equation, you have|− 16 |=| 16 |. Thus, the solution set
forxis{18,− 2 }.Example 3
Solve forx:| 11 − 3 x|= 1 −x.
Depending on the value of (11− 3 x) whether it is greater than or less than 0 the given
equation could be written as 11− 3 x= 1 −xor 11− 3 x=−(1−x). Solving both equations,
you havex=5 andx=3. Checkingx=5 with the original equation yields| 11 −3(5)|=
1 −5or|− 4 |=−4, which isnotpossible. Checkingx=3 with the original equation
shows that| 11 −3(3)|= 1 −3or| 2 |=−2 which is alsonotpossible. Thus, the solution for
xis the empty set{}. You could also solve the equation by using a graphing calculator.
Entery 1 =| 11 − 3 x|andy 2 = 1 −x. The two graphs do not intersect; thus, there is no com-
mon solution. (See Figure 5.2-1.)