#### Chapter 3 Inequalities in one Variable

Section 3.3 Absolute Value Inequalities and Quadratic Inequalities

# 3.3.1 Introduction

As in the approach in Module 2 and in the previous section, absolute values in inequalities are solved by a case analysis:
##### Info 3.3.1

To solve an absolute value inequality two cases are distinguished:
• For those values of $x$ for which the term beween absolute value bars is non-negative, the absolute value can be omitted or replaced by simple brackets, respectively.
• For those values of $x$ for which the term between absolute value bars is negative, the term is bracketed and negated.

Then, the solution sets arising from the case analysis will be restricted as described in the previous module and merged to form the solution set of the initial inequality.

##### Example 3.3.2
To solve the absolute value inequality $|4x-2|<1$, two cases are distinguished:
• For $x\ge \frac{1}{2}$, the term between absolute value bars is non-negative: In this case the inequality is equivalent to $\left(4x-2\right)<1$ or $x<\frac{3}{4}$, respectively. Because of the case condition the solution set is only $L{}_{1}=\left[\frac{1}{2};\frac{3}{4}\left[\text{}$ in this case.
• For $x<\frac{1}{2}$, the term between absolute value bars is negative: In this case the inequality is equivalent to $-\left(4x-2\right)<1$ or $x>\frac{1}{4}$, respectively. Only the subset $L{}_{2}=\text{}\right]\frac{1}{4};\frac{1}{2}\left[\text{}$ satisfies the case condition and contributes to the overall solution set.
The union of the two solution intervals results in the solution set $L=\text{}\right]\frac{1}{4};\frac{3}{4}\left[\text{}$ for the initial absolute value inequality: ##### Exercise 3.3.3
To solve the absolute value inequality $|x-1|<2|x-1|+x$ two cases are distinguished:
1. On the interval
, both terms in the absolute value terms are non-negative. The solution set of the inequality is in this case $L{}_{1}$$=$ .
2. On the interval
, both terms in the absolute value terms are negative. The solution set of the inequality is in this case $L{}_{2}$$=$ .
The union of the two intervals results in the solution interval $L$$=$
.