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In this section, we review some of the basic concepts involving the absolute value of a real number \(x\). There are a few
different ways to define absolute value and in this section we choose the following definition. (Absolute value
will be revisited in much greater depth in Section AbsoluteValueFunctions where we present what one can think of as the “precise”
definition.)
Absolute Value as Distance: For every real number \(x\), the absolute value of \(x\), denoted \(|x|\), is the distance between \(x\) and \(0\) on the
number line. More generally, if \(x\) and \(c\) are real numbers, \(|x-c|\) is the distance between the numbers \(x\) and \(c\) on the number
line.
For example, \(|5| = 5\) and \(|-5| = 5\), since each is \(5\) units from \(0\) on the number line:
Graphically why \(|-5|=5\) and \(|5|=5\).
Computationally, the absolute value ‘makes negative numbers positive’, though we need to be a little cautious with this
description. While \(|-7| = 7\), \(|5-7| \neq 5+7\). The absolute value acts as a grouping symbol, so \(|5-7| = |-2| = 2\), which makes sense since \(5\) and \(7\) are two units away
from each other on the number line:
Graphically why \(|5-7|=2\)
We list some of the operational properties of absolute value below.
Properties of Absolute Value: Let \(a\) and \(b\) be real numbers and let \(n\) be an integer. (See page ?? if you don’t remember
what an integer is.)
Product Rule: \(|ab|= |a||b|\)
Power Rule: \(\left | a^{n} \right | = |a|^{n}\) whenever \(a^{n}\) is defined
The proof of Theorem absolutevalueprops is difficult, but not impossible, using the distance definition of absolute value or even the ‘it makes
negatives positive’ notion. It is, however, much easier if one uses the “precise” definition given in Section AbsoluteValueFunctions so we will revisit
the proof then. For now, let’s focus on how to solve basic equations and inequalities involving the absolute
value.
1 Absolute Value Equations
Thinking of absolute value in terms of distance gives us a geometric way to interpret equations. For example, to solve \(|x| = 3\), we are
looking for all real numbers \(x\) whose distance from \(0\) is \(3\) units. If we move three units to the right of \(0\), we end up at \(x = 3\). If we move
three units to the left, we end up at \(x = -3\). Thus the solutions to \(|x| = 3\) are \(x = \pm 3\).
The solutions to \(|x| = 3\) are \(x=\pm 3\).
Thinking this way gives us the following.
Absolute Value Equations: Suppose \(x\), \(y\) and \(c\) are real numbers.
\(|x| = 0\) if and only if \(x = 0\).
For \(c > 0\), \(|x| = c\) if and only if \(x = c\) or \(x = -c\).
For \(c < 0\), \(|x| = c\) has no solution.
\(|x| = |y|\) if and only if \(x = y\) or \(x = -y\).
(That is, if two numbers have the same absolute values, they are either the same number or exact opposites
of each other.)
Theorem 2 is our main tool in solving equations involving the absolute value, since it allows us a way to rewrite such equations
as compound linear equations.
Strategy for Solving Equations Involving Absolute Value
In order to solve an equation involving the absolute value of a quantity \(|X|\):
Isolate the absolute value on one side of the equation so it has the form \(|X| = c\).
The techniques we use to ‘isolate the absolute value’ are precisely those we used in Section ?? to isolate the variable when
solving linear equations. Time for some practice.
Solve each of the following equations.
\(|3x-1| = 6\)
\(\dfrac {3 - |y+5|}{2} = 1\)
\(3|2t+1| - \sqrt {5} = 0\)
\(4 - |5w+3| = 5\)
\(\left |3 - x \sqrt [3]{12}\right | = |4x+1|\)
\(|t-1| - 3|t+1| = 0\)
Solution.
The equation \(|3x-1| = 6\) is of already in the form \(|X| = c\), so we know that either \(3x-1=6\) or \(3x-1 = -6\). Solving the former gives us at \(x = \frac {7}{3}\) and solving
the latter yields \(x = -\frac {5}{3}\). We may check both of these solutions by substituting them into the original equation and
showing that the arithmetic works out.
We begin solving \(\frac {3 - |y+5|}{2} = 1\) by isolating the absolute value to put it in the form \(|X| = c\).
At this point, we have \(y+5 = 1\) or \(y+5 = -1\), so our
solutions are \(y = -4\) or \(y = -6\). We leave it to the reader to check both answers in the original equation.
As in the previous example, we first isolate the absolute value. Don’t let the \(\sqrt {5}\) throw you off - it’s just another
real number, so we treat it as such:
From here, we have that \(2t+1 = \frac {\sqrt {5}}{3}\) or \(2t+1 = -\frac {\sqrt {5}}{3}\). The first equation gives \(t = \frac {\sqrt {5}-3}{6}\) while the second
gives \(t = \frac {-\sqrt {5}-3}{6}\) thus we list our answers as \(t = \frac {-3 \pm \sqrt {5}}{6}\). The reader should enjoy the challenge of substituting both answers into the
original equation and following through the arithmetic to see that both answers work.
Upon isolating the absolute value in the equation \(4 - |5w+3| = 5\), we get \(|5w+3| = -1\). At this point, we know there cannot be any real
solution. By definition, the absolute value is a distance, and as such is never negative. We write ‘no solution’
and carry on.
Our next equation already has the absolute value expressions (plural) isolated, so we work from the principle
that if \(|x| = |y|\), then \(x = y\) or \(x = -y\). Thus from \(\left |3 - x \sqrt [3]{12}\right | = |4x+1|\) we get two equations to solve:
\[ 3 - x \sqrt [3]{12} = 4x+1, \qquad \text {and} \qquad 3 - x \sqrt [3]{12} = -(4x+1) \]
Notice that the right side of the second equation is
\(-(4x+1)\) and not simply \(-4x+1\). Remember, the expression \(4x+1\) represents a single real number so in order to negate it we need
to negate the entire expression \(-(4x+1)\). Moving along, when solving \(3 - x \sqrt [3]{12} = 4x+1\), we obtain \(x = \frac {2}{4 + \sqrt [3]{12}}\) and the solution to \(3 - x \sqrt [3]{12} = -(4x+1)\) is \(x = \frac {4}{\sqrt [3]{12}-4}\). As usual, the
reader is invited to check these answers by substituting them into the original equation.
We start by isolating one of the absolute value expressions: \(|t-1| - 3|t+1| = 0\) gives \(|t-1| = 3|t+1|\). While this resembles the form \(|x| = |y|\), the
coefficient \(3\) in \(3|t+1|\) prevents it from being an exact match. Not to worry - since \(3\) is positive, \(3 = |3|\) so
\[3|t+1| = |3| |t+1| = |3(t+1)| = |3t+3|.\]
Hence, our equation
becomes \(|t-1| = |3t+3|\) which results in the two equations: \(t-1 = 3t+3\) and \(t-1 = -(3t+3)\). The first equation gives \(t = -2\) and the second gives \(t = -\frac {1}{2}\). The reader
is encouraged to check both answers in the original equation. □
2 Absolute Value Inequalities
We now turn our attention to solving some basic inequalities involving the absolute value. Suppose we wished to solve \(|x| < 3\).
Geometrically, we are looking for all of the real numbers whose distance from \(0\) is less than \(3\) units. We get \(-3 < x < 3\), or in interval
notation, \((-3,3)\). Suppose we are asked to solve \(|x| > 3\) instead. Now we want the distance between \(x\) and \(0\) to be greater than \(3\) units. Moving
in the positive direction, this means \(x > 3\). In the negative direction, this puts \(x < -3\). Our solutions would then satisfy \(x < -3\)or\(x > 3\). In interval
notation, we express this as \((-\infty , -3) \cup (3, \infty )\).
The solution to \(|x|<3\) is \((-3,3)\)
The solution to \(|x|>3\) is \((-\infty ,-3)\cup (3,\infty )\)
Generalizing this notion, we get the following:
Inequalities Involving Absolute Value: Let \(c\) be a real number.
If \(c> 0\), \(|x| < c\) is equivalent to \(-c<x<c\).
If \(c \leq 0\), \(|x| < c\) has no solution.
If \(c > 0\), \(|x| > c\) is equivalent to \(x < -c\) or \(x > c\).
If \(c \leq 0\), \(|x| > c\) is true for all real numbers.
If the inequality we’re faced with involves ‘\(\leq \)’ or ‘\(\geq \),’ we can combine the results of Theorem 3 with Theorem 2 as
needed.
Strategy for Solving Inequalities Involving Absolute Value
In order to solve an inequality involving the absolute value of a quantity \(|X|\):
Isolate the absolute value on one side of the inequality.
From Theorem 3, \(\left |x-\sqrt [4]{5} \right | > 1\) is equivalent to \(x-\sqrt [4]{5} < -1\) or \(x-\sqrt [4]{5} > 1\). Solving this compound inequality, we get \(x < -1 + \sqrt [4]{5}\) or \(x > 1 + \sqrt [4]{5}\). Our answer, in interval
notation, is: \(\left (-\infty ,-1 + \sqrt [4]{5} \right ) \cup \left (1 + \sqrt [4]{5},\infty \right )\). As with linear inequalities, we can only partially check our answer by selecting values of \(x\) both
inside and outside of the solution intervals to see which values of \(x\) satisfy the original inequality and which do
not.
Our first step in solving \(\frac {4 - 2|2x+1|}{4} \geq -\sqrt {3}\) is to isolate the absolute value.
Since we’re dealing with ‘\(\leq \)’ instead of just ‘\(<\),’ we can
combine Theorems 3 and 2 to rewrite this last inequality as: (Note the use of parentheses: \(-(2+2\sqrt {3})\) as opposed
to \(-2 + 2\sqrt {3}\).)\(-(2 + 2\sqrt {3}) \leq 2x+1 \leq 2+2\sqrt {3}\). Subtracting the ‘\(1\)’ across both inequalities gives \(-3-2\sqrt {3} \leq 2x \leq 1 + 2\sqrt {3}\), which reduces to \(\frac {-3-2\sqrt {3}}{2} \leq x \leq \frac {1+2\sqrt {3}}{2}\). In interval notation this reads as \(\left [\frac {-3-2\sqrt {3}}{2}, \frac {1+2\sqrt {3}}{2}\right ]\).
There are two absolute values in \(|2x - 1| \leq 3|4 - 8x| - 10\), so we cannot directly apply Theorem 3 here. Notice, however, that \(|4 - 8x| = |(-4)(2x-1)|\). Using
this, we get:
Now we are allowed to invoke Theorems 2 and 3 and write the equivalent compound inequality: \(2x - 1 \leq -\frac {10}{11}\)or\(2x-1 \geq \frac {10}{11}\). We get \(x \leq \frac {1}{22}\)or\(x \geq \frac {21}{22}\), which when written with interval notation becomes \(\left (-\infty , \frac {1}{22}\right ] \cup \left [\frac {21}{22}, \infty \right )\).
The inequality \(|2x - 1| \leq 3|4 - 8x| + 10\) differs from the previous example in exactly one respect: on the right side of the inequality, we
have ‘\(+10\)’ instead of ‘\(-10\).’ The steps to isolate the absolute value here are identical to those in the previous example,
but instead of obtaining \(|2x - 1| \geq \frac {10}{11}\) as before, we obtain \(|2x - 1| \geq -\frac {10}{11}\). This latter inequality is always true. (Absolute value is, by
definition, a distance and hence always \(0\) or greater.) Thus our solution to this inequality is all real numbers.
To solve \(2 < |x-1| \leq 5\), we rewrite it as the compound inequality: \(2 < |x-1|\)and\(|x-1| \leq 5\). The first inequality, \(2 < |x-1|\), can be re-written as \(|x-1|>2\) so it is
equivalent to \(x-1 < -2\)or\(x-1 > 2\). Thus the solution to \(2 < |x-1|\) is \(x<-1\) or \(x>3\), which in interval notation is \((-\infty , -1) \cup (3, \infty )\). For \(|x-1| \leq 5\), we combine the results of
Theorems 2 and 3 to get \(-5 \leq x-1 \leq 5\) so that \(-4 \leq x \leq 6\), or \([-4,6]\).
Our solution to \(2 < |x-1| \leq 5\) is comprised of values of \(x\) which satisfy both parts of the inequality, so we intersect \((-\infty , -1) \cup (3, \infty )\) with \([-4,6]\) to get
our final answer \([-4,-1) \cup (3,6]\).
Our first hope when encountering \(|10 x - 5| + |10 - 5x| \leq 0\) is that we can somehow combine the two absolute value quantities as we’d
done in earlier examples. We leave it to the reader to show, however, that no matter what we try to factor out
of the absolute value quantities, what remains inside the absolute values will always be different.
At this point, we take a step back and look at the equation in a more general way: we are adding two absolute
values together and wanting the result to be less than or equal to \(0\). The absolute value of anything is always \(0\)
or greater, so there are no solutions to: \(|10x - 5| + |10 - 5x| < 0\).
Is it possible that \(|10x - 5| + |10 - 5x| = 0\)? Only if there is an \(x\) where \(|10x-5| = 0\) and \(|10-5x| = 0\)at the same time. (Do you see why?) The first equation
holds only when \(x = \frac {1}{2}\), while the second holds only when \(x = 2\). Alas, we have no solution. (Not for lack of trying,
however!) □
The astute reader will have noticed by now that the authors have done nothing in the way of explaining why anyone would
ever need to know this stuff. Go back and read the New Preface and the introduction to the Appendix. These sections are
designed to review skills and concepts that you’ve already learned. Thus the deeper applications are in the main body of the
text as opposed to here in the Appendix.
We close this section with an example of how the properties in Theorem 1 are used in Calculus. Here, ‘\(\varepsilon \)’ is the Greek letter
‘epsilon’ and it represents a positive real number. Those of you who will be taking Calculus in the future should become very
familiar with this type of algebraic manipulation.
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