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1.2: The absolute value

  • Page ID
    48948
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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 \(S=\{-7,-1\}\).


    This page titled 1.2: The absolute value is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Thomas Tradler and Holly Carley (New York City College of Technology at CUNY Academic Works) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.