1.2: The absolute value
- Page ID
- 48948
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\dsum}{\displaystyle\sum\limits} \)
\( \newcommand{\dint}{\displaystyle\int\limits} \)
\( \newcommand{\dlim}{\displaystyle\lim\limits} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\(\newcommand{\longvect}{\overrightarrow}\)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\ket}[1]{\left| #1 \right>}\)
\(\newcommand{\bra}[1]{\left< #1 \right|}\)
\(\newcommand{\braket}[2]{\left< #1 \vphantom{#2} \right| \left. #2 \vphantom{#1} \right>}\)
\(\newcommand{\braopket}[3]{\left< #1 \vphantom{#2}\vphantom{#3} \right| #2 \vphantom{#1}\vphantom{#3} \left| #3 \vphantom{#1}\vphantom{#2} \right>}\)
\(\newcommand{\qmvec}[1]{\mathbf{\vec{#1}}}\)
\(\newcommand{\op}[1]{\hat{\mathbf{#1}}}\)
\(\newcommand{\expect}[1]{\langle #1 \rangle}\)
\(\newcommand{\dfn}[1]{\emph{\textbf{#1}}}\)
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:
\[|c|=\left\{\begin{array}{cl}
c, & \text { if } c \text { is positive or zero } \\
-c, & \text { if } c \text { is negative }
\end{array}\right. \nonumber \]
\(|-4|=4\)
\(|12|=12\)
\(|-3.523|=3.523\)
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 \(S=\{-3,3\}\).
|x|=5] Solve for \(x\): \(|x|=5\)
Solution
\(x=5\) or \(x=-5\). The solution set is \(S=\{-5,5\}\).
|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 \(S=\{\}\).
Solve for \(x\): \(|x|=0\).
Solution
Since \(-0=0\), there is only one solution, \(x=0\). Thus, \(S=\{0\}\).
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 \(S=\{-8,4\}\).
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 \(S=\{-\frac 1 3,3\}\).
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 \(S=\{-7,-1\}\).