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1.2: The absolute value

Last updated

May 2, 2022
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- 1.1: Background regarding numbers
- 1.3: Inequalities and intervals

Thomas Tradler and Holly Carley
CUNY New York City College of Technology via New York City College of Technology at CUNY Academic Works

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The absolute value of a real number $c$ , denoted by $|c|$ the non-negative number which is equal in magnitude (or size) to $c$ , i.e., is the number resulting from disregarding the sign:

$|c|=\left\{\begin{array}{cl} c, & \text { if } c \text { is positive or zero } \\ -c, & \text { if } c \text { is negative } \end{array}\right. \nonumber$

Example $\PageIndex{1}$

$|-4|=4$

Example $\PageIndex{2}$

$|12|=12$

Example $\PageIndex{3}$

$|-3.523|=3.523$

Example $\PageIndex{4}$

For which real numbers $x$ do you have $|x|=3$ ?

Solution

Since $|3|=3$ and $|-3|=3$ , we see that there are two solutions, $x=3$ or $x=-3$ .

The solution set is $S=\{-3,3\}$ .

Example $\PageIndex{5}$

|x|=5] Solve for $x$ : $|x|=5$

Solution

$x=5$ or $x=-5$ . The solution set is $S=\{-5,5\}$ .

Example $\PageIndex{6}$

|x|=-7] Solve for $x$ : $|x|=-7$ .

Solution

Note that $|-7|=7$ and $|7|=7$ so that these cannot give any solutions. Indeed, there are no solutions, since the absolute value is always non-negative. The solution set is the empty set $S=\{\}$ .

Example $\PageIndex{7}$

Solve for $x$ : $|x|=0$ .

Solution

Since $-0=0$ , there is only one solution, $x=0$ . Thus, $S=\{0\}$ .

Example $\PageIndex{8}$

Solve for $x$ : $|x+2|=6$ .

Solution

Since the absolute value of $x+2$ is $6$ , we see that $x+2$ has to be either $6$ or $-6$ .

We evaluate each case,

$\begin{array}{l|l} \text { either } x+2=6, & \text { or } x+2=-6 \\ \Longrightarrow x=6-2, & \Longrightarrow x=-6-2 \\ \Longrightarrow x=4, & \Longrightarrow x=-8 \end{array} \nonumber$

The solution set is $S=\{-8,4\}$ .

Example $\PageIndex{9}$

Solve for $x$ : $|3x-4|=5$

Solution

$\begin{array}{l|l} \text { Either } 3 x-4=5 & \text { or } 3 x-4=-5 \\ \Longrightarrow 3 x=9 & \Longrightarrow 3 x=-1 \\ \Longrightarrow x=3 & \Longrightarrow x=-\frac{1}{3} \end{array} \nonumber$

The solution set is $S=\{-\frac 1 3,3\}$ .

Example $\PageIndex{10}$

Solve for $x$ : $-2\cdot |12+3x|=-18$

Solution

Dividing both sides by $-2$ gives $|12+3x|=9$ . With this, we have the two cases

$\begin{array}{l|l} \text { Either } 12+3 x=9 & \text { or } 12+3 x=-9 \\ \Longrightarrow 3 x=-3 & \Longrightarrow 3 x=-21 \\ \Longrightarrow x=-1 & \Longrightarrow x=-7 \end{array} \nonumber$

The solution set is $S=\{-7,-1\}$ .

Example 1.2.1\PageIndex{1}

Example 1.2.2\PageIndex{2}

Example 1.2.3\PageIndex{3}

Example 1.2.4\PageIndex{4}

Example 1.2.5\PageIndex{5}

Example 1.2.6\PageIndex{6}

Example 1.2.7\PageIndex{7}

Example 1.2.8\PageIndex{8}

Example 1.2.9\PageIndex{9}

Example 1.2.10\PageIndex{10}

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Example $\PageIndex{1}$

Example $\PageIndex{2}$

Example $\PageIndex{3}$

Example $\PageIndex{4}$

Example $\PageIndex{5}$

Example $\PageIndex{6}$

Example $\PageIndex{7}$

Example $\PageIndex{8}$

Example $\PageIndex{9}$

Example $\PageIndex{10}$