Absolute value equations and inequalities

Overview

Absolute value represents the distance of a number from zero on the number line, always non-negative. Solving equations and inequalities involving absolute values requires understanding this geometric meaning and translating it into algebraic cases.

For inequalities like $|x| < 5$, we ask: "What numbers are less than 5 units from zero?" Answer: everything between $-5$ and $5$. For $|x| > 5$: "What's more than 5 units away?" Answer: everything to the left of $-5$ OR to the right of $5$.

Definition and Foundation

WHY this definition? We want a function that measures distance, which is always positive. For positive numbers, the distance from zero IS the number itself. For negative numbers, we flip the sign to make it positive.

Key property: $|x| = \sqrt{x^2}$ for all real $x$.

WHY? Squaring any number makes it positive, then taking the square root gives the positive root. This connects absolute value to the Pythagorean distance concept.

Absolute Value Equations

Case Analysis Method

To solve $|A| = b$ where $b \geq 0$:

Derivation from first principles:

1. Start with the definition: $|A|$ equals $A$ when $A \geq 0$, and equals $-A$ when $A < 0$
2. So $|A| = b$ means either:
   • Case 1: $A = b$ (when the expression inside is positive or zero)
   • Case 2: $-A = b$, which gives $A = -b$ (when the expression inside is negative)

WHY two cases? The absolute value "strips away" the sign. Both $5$ and $-5$ become $5$ after applying absolute value. Working backward, if the result is $5$, the original could have been either.

Special case: If $b < 0$, there is no solution because absolute values are never negative.

Solution:

• Case 1: $x - 3 = 7 \implies x = 10$
  • Why this step? We assume the inside is positive, so absolute value doesn't change it
• Case 2: $x - 3 = -7 \implies x = -4$
  • Why this step? We assume the inside is negative, so absolute value flips its sign (hence we set it equal to the negative of 7)

Verification:

• $|10 - 3| = |7| = 7$ ✓
• $|-4 - 3| = |-7| = 7$ ✓

Geometric interpretation: We found all numbers that are exactly 7 units away from 3 on the number line: $3 + 7 = 10$ and $3 - 7 = -4$.

Solution: First, check when the right side is non-negative: we need $3x - 1 \geq 0 \implies x \geq \frac{1}{3}$

• Case 1: $2x + 5 = 3x - 1$

  • Why? Assume $2x + 5 \geq 0$, so absolute value leaves it unchanged
  • $5 + 1 = 3x - 2x$
  • $x = 6$
  • Check the sign assumption: $2(6) + 5 = 17 \geq 0$ ✓ (inside is indeed non-negative)
  • Check RHS non-negative: $x = 6 \geq \frac{1}{3}$ ✓
  • Verify: $|2(6) + 5| = 17$ and $3(6) - 1 = 17$ ✓

• Case 2: $2x + 5 = -(3x - 1)$

  • Why? Assume $2x + 5 < 0$, so absolute value flips the sign
  • $2x + 5 = -3x + 1$
  • $5x = -4$
  • $x = -\frac{4}{5}$
  • Check the sign assumption: $2(-\frac{4}{5}) + 5 = -\frac{8}{5} + 5 = \frac{17}{5} > 0$ ✗ — but we assumed $2x + 5 < 0$! The assumption is violated, so this branch is invalid here.
  • Check RHS non-negative: $x = -\frac{4}{5} < \frac{1}{3}$ ✗ ($3x - 1 = -\frac{17}{5} < 0$)
  • This solution is extraneous on both counts!

Answer: $x = 6$ only

Why did Case 2 fail? Two independent reasons: (1) the sign assumption "$2x + 5 < 0$" is contradicted since $2(-\frac{4}{5})+5 = \frac{17}{5} > 0$, and (2) the right side $3x - 1 = -\frac{17}{5}$ is negative, but absolute values can never equal a negative number. Always check both the sign assumption of the inner expression AND the sign of the RHS.

Absolute Value Inequalities

Less Than Inequalities

To solve $|A| < b$ where $b > 0$:

Derivation:

• $|A| < b$ means "the distance from $A$ to zero is less than $b$"
• This happens when $A$ is between $-b$ and $b$
• Algebraically: $-b < A < b$

WHY? If $A \geq 0$: $|A| = A < b$. If $A < 0$: $|A| = -A < b \implies A > -b$. Combining both cases: $-b < A < b$.

Solution: $-5 \leq x + 2 \leq 5$

Why this step? We're finding all $x$ where $x + 2$ is at most 5 units from zero, which means between $-5$ and $5$.

Subtract 2 from all parts: $-7 \leq x \leq 3$

Interval notation: $[-7, 3]$

Geometric check: The center is at $x = -2$ (where $x + 2 = 0$). We go 5 units in each direction: $-2 - 5 = -7$ and $-2 + 5 = 3$. ✓

Greater Than Inequalities

To solve $|A| > b$ where $b > 0$:

Derivation:

• $|A| > b$ means "the distance from $A$ to zero is greater than $b$"
• This happens when $A$ is outside the interval $[-b, b]$
• So either $A > b$ (far right) OR $A < -b$ (far left)

WHY the OR? The number line has two regions far from zero: the positive side and the negative side. Being more than $b$ units away means being in either region, not both simultaneously.

Solution: Split into two cases:

• Case 1: $3x - 1 > 8$

  • Why? The expression is far to the right
  • $3x > 9$
  • $x > 3$

• Case 2: $3x - 1 < -8$

  • Why? The expression is far to the left
  • $3x < -7$
  • $x < -\frac{7}{3}$

Answer: $x \in (-\infty, -\frac{7}{3}) \cup (3, \infty)$

Geometric interpretation: We want $x$ values where $3x - 1$ is more than 8 units from zero. The center is at $3x - 1 = 0\implies x = \frac{1}{3}$. Everything within $\frac{8}{3}$ units of $\frac{1}{3}$ is excluded, everything outside is included.

Advanced Techniques

Compound Absolute Value Equations

Solution: Identify critical points where expressions inside change sign: $x = 1$ and $x = -2$. These divide the number line into three regions.

Why critical points? At these points, the expressions equal zero, and the absolute value behavior switches.

• Region 1: $x < -2$ (both expressions negative)

  • $|x - 1| = -(x - 1) = 1 - x$
  • $|x + 2| = -(x + 2) = -x - 2$
  • Equation: $(1 - x) + (-x - 2) = 5$
  • $-2x - 1 = 5 \implies x = -3$
  • Check: $x = -3 < -2$ ✓
  • Verify: $|-3 - 1| + |-3 + 2| = 4 + 1 = 5$ ✓

• Region 2: $-2 \leq x < 1$ (first negative, second positive)

  • $|x - 1| = 1 - x$
  • $|x + 2| = x + 2$
  • Equation: $(1 - x) + (x + 2) = 5$
  • $3 = 5$ ✗
  • No solution in this region!

• Region 3: $x \geq 1$ (both positive)

  • $|x - 1| = x - 1$
  • $|x + 2| = x + 2$
  • Equation: $(x - 1) + (x + 2) = 5$
  • $2x + 1 = 5 \implies x = 2$
  • Check: $x = 2 \geq 1$ ✓
  • Verify: $|2 - 1| + |2 + 2| = 1 + 4 = 5$ ✓

Answer: $x \in \{-3, 2\}$

Common Mistakes and Fixes

Why it feels right: We see an equation and solve it directly, forgetting absolute value has two "branches."

The fix: Always write both cases: $x - 2 = 5$ AND $x - 2 = -5$. The geometric intuition helps: "What's 5 units from 2?" has two answers: $2 + 5 = 7$ and $2 - 5 = -3$.

Why it feels right: Students memorize "$|x| < 3$ means $-3 < x < 3$" and try to apply the same pattern.

The fix: Greater-than means outside the interval, so use OR: $x > 3$ or $x < -3$. Remember: "less than" is AND (between), "greater than" is OR (outside). The union symbol $\cup$ interval notation reminds us it's separate regions.

Why it feels right: We trust our algebra and assume both case solutions are valid.

The fix: When the equation is $|A| = B$ with $B$ containing variables, always verify that $B \geq 0$ for your solution, AND that the sign assumption you made about $A$ holds. If $B < 0$ at your $x$-value, that solution is extraneous. Example 2 above shows both checks in action.

Why it feels right: Students confuse the "split into cases" approach from equations with the inequality rules.

The fix: For $|A| < b$, write the single compound inequality $-b < A < b$. Don't split! For $|A| > b$, then you split into $A > b$ OR $A < -b$. Pattern: "less than" stays together (AND), "greater than" splits apart (OR).

Memory Aids

• $|x| < 5$: LANDS between $-5$ and $5$ → $-5 < x < 5$ (one interval, AND)
• $|x| > 5$: GOES to extremes → $x < -5$ or $x > 5$ (two intervals, OR)

Visual: Draw a number line. "Less than" shades the middle (landing zone). "Greater than" shades the ends (extremes).

Active Recall Practice

#flashcards/maths

What does $|x| = a$ translate to in terms of cases? :: Two equations: $x = a$ or $x = -a$ (provided $a \geq 0$). Geometric: numbers exactly $a$ units from zero.

When we solve $|x| = 3$, we're asking: "Where can I stand to be exactly 3 steps from zero?" Two places: at 3 (three steps right) and at -3 (three steps left).

For $|x| < 3$, we want: "Where can I stand to be fewer than 3 steps from zero?" Answer: anywhere between -3 and 3, because those are the only spots close enough.

For $|x| > 3$, we want: "Where can I stand to be more than 3 steps from zero?" Answer: either past 3 on the right, OR past -3 on the left — the far away places on both sides!

The tricky part: when the problem is something like $|x - 2| = 5$, now zero has moved! We're measuring 5 steps from 2 instead of 0. So our answers are $2 + 5 = 7$ and $2 - 5 = -3$.

Connections

• Algebraic equations — absolute value equations are special cases requiring analysis
• Linear inequalities — absolute value inequalities combine linear inequalities with OR/AND logic
• Number line and intervals — geometric interpretation of absolute value as distance
• Function transformations — $|f(x)|$ reflects negative parts of $f(x)$ upward
• Piecewise functions — absolute value is defined piecewise: $|x| = x$ if $x \geq 0$, $-x$ if $x < 0$
• Distance formula — $|x - a|$ is the distance between $x$ and $a$ on the number line, generalizes to $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ in 2D
• Quadratic equations — squaring both sides of $|A| = b$ gives $A^2 = b^2$, another solution method

Master absolute value by always thinking: "distance from zero" for the geometric intuition, then translate systematically into cases.

Concept Map

Hinglish (regional understanding)

Hinglish (regional understanding)