Principle 4.14. Absolute Value Equations.
If \(|x| = A\text{,}\) where \(A \gt 0\text{,}\) then \(x = \pm A\text{.}\) If \(|x|=0\text{,}\) then \(x=0\text{.}\) Finally, the equation \(|x| = -A\) has no solution since \(|x|\geq 0\text{.}\)
Section 4.3 Absolute Value Equations
Recall that \(|x|\text{,}\) the absolute value of \(x\) (1.7), measures how far \(x\) is from zero. So the equation
\begin{equation*}
|x| = 4
\end{equation*}
tells us that \(x\) must be within \(4\) units of zero. There are only two possibilities for \(x\text{:}\)
\begin{equation*}
x = 4 \quad \text{OR} \quad x = -4
\end{equation*}
which we write compactly as \(x=\pm 4\text{.}\) This method is summarized by the following principle.
Example 4.15. Solving a Linear Absolute Value Equation.
Let’s solve
\begin{equation*}
|2x-3| +6 = 10.
\end{equation*}
First, isolate the term with the absolute value.
\begin{equation*}
|2x-3 | = 4
\end{equation*}
Thus, the number \(2x-3\) must be within \(4\) units of zero. There are only two possibilities
\begin{equation*}
2x-3 = 4 \quad \text{or} \quad 2x-3 = -4
\end{equation*}
Solving the first,
\begin{equation*}
2x-3 = 4 \Rightarrow 2x = 7 \Rightarrow x = 7/2
\end{equation*}
Solving the second,
\begin{equation*}
2x-3 = -4 \Rightarrow 2x = -1 \Rightarrow x = -1/2
\end{equation*}
Thus, the solutions to our equation are
\begin{equation*}
x = 7/2, \, -1/2.
\end{equation*}
Inequalities involving absolute values are useful in Calculus. For example, the inequality \(|x| \lt 4\) says that \(x\) must be within four units of zero. This requires \(x\) satisfy the compound inequality
\begin{equation*}
-4 \lt x \quad \text{AND} \quad x \lt 4,
\end{equation*}
or equivalently, \(-4 \lt x \lt 4\text{.}\) Similarly, the inequality \(|x| \gt 4\) requires that \(x\) be further than \(4\) units from zero. This requires \(x\) satisfy the compound inequality
\begin{equation*}
x \lt -4 \quad \text{OR}\quad 4 \lt x.
\end{equation*}
Principle 4.16. Absolute Value Inequalities.
- If \(|x| \lt A\text{,}\) where \(A \gt 0\text{,}\) then\begin{equation*} -A \lt x \lt A\text{.} \end{equation*}
- If \(|x| \gt A\text{,}\) where \(A \gt 0\text{,}\) then\begin{equation*} x\lt -A \quad \text{OR}\quad A \lt x. \end{equation*}
Example 4.17. Solving an Absolute Value Inequality.
Let’s solve
\begin{equation*}
\left|2-3x\right| \leq 10.
\end{equation*}
We require that \(2-3x\) is within \(10\) units of zero. So
\begin{gather*}
-10 \leq 2-3x \leq 10\\
-12 \leq -3x \leq 8\\
-4 \leq -x \leq \frac{8}{3}\\
4 \geq x \geq -\frac{8}{3}\\
-\frac{8}{3} \leq x \leq 4
\end{gather*}
So the solution is the interval \(\left[-\frac{8}{3},4\right]\text{.}\)
Example 4.18. Solving an Absolute Value Inequality.
Solve
\begin{equation*}
\left|x-7\right| -5 \gt 6.
\end{equation*}
Solution.
First, isolate the absolute value to obtain
\begin{equation*}
\left|x-7\right| \gt 11.
\end{equation*}
We require that \(x-7\) is more than \(11\) units from zero. So
\begin{gather*}
x-7 \lt -11 \quad \text{OR} \quad 11 \lt x-7\\
x\lt -4 \quad \text{OR} \quad 18 \lt x
\end{gather*}
So the solution is the union \((-\infty,-4)\cup(18,+\infty)\text{.}\)
