1.2: The absolute value

Last updated
Save as PDF

Page ID: 48948

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

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

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 \(S=\{-7,-1\}\).