3.4 Absolute Value Equations & Absolute Value Inequalities

3.4 Absolute Value Equations & Absolute Value Inequalities

The absolute value of a real number $x$, written $|x|$, represents its magnitude or distance from zero on the real number line, regardless of direction. Because distance is non-negative, $|x| \ge 0$ for all real numbers $x$.

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1. Geometric Interpretation & Piecewise Definition

Formally, absolute value is defined as a two-piece piecewise function:

$|x| = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{cases}$

Distance on the Real Line

Geometrically, the equation $|x - a| = c$ (where $c \ge 0$) signifies: "The distance between variable $x$ and fixed point $a$ on the number line is $c$ units." Point $a$ acts as the center and $c$ acts as the distance (or tolerance).

Real-World Application: Manufacturing Tolerance

In industrial quality assurance, absolute value inequalities specify acceptable engineering tolerances. If a machine fills beverage bottles to a target volume of $16.0\text{ oz}$ with an allowable tolerance of $0.2\text{ oz}$, volume $x$ satisfies:

$|x - 16.0| \le 0.2 \implies -0.2 \le x - 16.0 \le 0.2 \implies 15.8 \le x \le 16.2$

Bottles containing between $15.8\text{ oz}$ and $16.2\text{ oz}$ pass inspection.

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2. Solving Absolute Value Equations ($|u| = c$)

To solve $|u| = c$:

1. If $c > 0$: Split into two linear equations: $u = c$ or $u = -c$.
2. If $c = 0$: Solve the single linear equation $u = 0$.
3. If $c < 0$: No solution ($\emptyset$), because absolute value cannot equal a negative number.

Worked Example 1: Standard Absolute Value Equation

Solve $|2x - 7| = 9$.

1. Split into two cases:
   • Case 1: $2x - 7 = 9 \implies 2x = 16 \implies x = 8$
   • Case 2: $2x - 7 = -9 \implies 2x = -2 \implies x = -1$
2. Check: $|2(8) - 7| = |9| = 9$, and $|2(-1) - 7| = |-9| = 9$. Solution set: $\{-1, 8\}$.

Equations with Two Absolute Values ($|u| = |v|$)

When $|u| = |v|$, expressions $u$ and $v$ must have equal magnitude, meaning they are either identical or opposite in sign: $u = v$ or $u = -v$.

Worked Example 2: Double Absolute Value

Solve $|3x - 1| = |x + 5|$.

• Case 1: $3x - 1 = x + 5 \implies 2x = 6 \implies x = 3$
• Case 2: $3x - 1 = -(x + 5) \implies 3x - 1 = -x - 5 \implies 4x = -4 \implies x = -1$ Solution set: $\{-1, 3\}$.

Equations with Variable Expressions on the Right ($|u| = g(x)$)

When solving $|ax + b| = cx + d$, the right side $cx + d$ must be non-negative ($cx + d \ge 0$). Candidate roots causing $cx + d < 0$ are extraneous solutions and must be eliminated.

Worked Example 3: Extraneous Solution Check

Solve $|3x + 2| = x - 4$.

1. Split into cases:
   • Case 1: $3x + 2 = x - 4 \implies 2x = -6 \implies x = -3$
   • Case 2: $3x + 2 = -(x - 4) \implies 3x + 2 = -x + 4 \implies 4x = 2 \implies x = \frac{1}{2}$
2. Screen Candidate Roots:
   • Check $x = -3$: Left side = $|3(-3) + 2| = |-7| = 7$. Right side = $-3 - 4 = -7$. Since $7 \neq -7$, $x = -3$ is extraneous.
   • Check $x = \frac{1}{2}$: Left side = $\left|3\left(\frac{1}{2}\right) + 2\right| = \frac{7}{2}$. Right side = $\frac{1}{2} - 4 = -\frac{7}{2}$. Since $\frac{7}{2} \neq -\frac{7}{2}$, $x = \frac{1}{2}$ is extraneous.
3. Conclusion: No solution ($\emptyset$).

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3. Absolute Value Inequalities ($|u| < c$ vs. $|u| > c$)

Less-Than Inequalities: Conjunctions (AND)

For $c > 0$, $|u| < c$ bounds expression $u$ between $-c$ and $c$:

$-c < u < c$

Worked Example 4: Less-Than Inequality

Solve $|2x - 5| \le 9$ and write in interval notation.

1. Translate to bounded compound inequality: $-9 \le 2x - 5 \le 9$
2. Add $5$ to all three parts: $-4 \le 2x \le 14$
3. Divide all three parts by $2$: $-2 \le x \le 7$ Interval notation: $[-2, 7]$.

Greater-Than Inequalities: Disjunctions (OR)

For $c > 0$, $|u| > c$ splits into two outer rays:

$u < -c \quad \text{or} \quad u > c$

Worked Example 5: Greater-Than Inequality

Solve $|3 - 4x| > 11$.

1. Split into two separate inequalities:
   • Left: $3 - 4x < -11 \implies -4x < -14 \implies x > \frac{7}{2}$
   • Right: $3 - 4x > 11 \implies -4x > 8 \implies x < -2$
2. Combine using Union ($\cup$): Interval notation: $(-\infty, -2) \cup \left(\frac{7}{2}, \infty\right)$.

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4. Degenerate & Special Edge Cases Matrix Table