1.2: The absolute value
- Page ID
- 48948
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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 \(S=\{-7,-1\}\).