2.15 Modulus Equations and Inequalities (HL)
1. Core Principles of Modulus
The mathematical operation defining absolute value establishes precise boundary relationships relative to a positive structural constant $a$:
- $|x| = a \Leftrightarrow x = a \text{ or } x = -a$
- $|x| < a \Leftrightarrow -a < x < a$ (Defines an interior bounded space)
- $|x| > a \Leftrightarrow x < -a \text{ or } x > a$ (Defines an exterior divergent space)
EXAMPLE 1
$2x - 3 = 5 \text{ or } 2x - 3 = -5 \Rightarrow 2x = 8 \text{ or } 2x = -2 \Rightarrow \mathbf{x = 4 \text{ or } x = -1}$.
$-5 < 2x - 3 < 5 \Rightarrow -2 < 2x < 8 \Rightarrow \mathbf{-1 < x < 4}$.
$2x - 3 < -5 \text{ or } 2x - 3 > 5 \Rightarrow 2x < -2 \text{ or } 2x > 8 \Rightarrow \mathbf{x < -1 \text{ or } x > 4}$.
2. Modulus with Variable Boundaries
The application of standard bounds is heavily disrupted when the boundary equation involves uncontrolled mathematical variables.
EXAMPLE 2
Isolate both geometric states:
$x - 1 = 5x - 10 \text{ or } x - 1 = -5x + 10$
$4x = 9 \text{ or } 6x = 11 \Rightarrow \mathbf{x = 9/4 \text{ or } x = 11/6}$.
Because variables introduce moving constraints, the domain interval $-5x + 10 < x - 1 < 5x - 10$ must be calculated structurally across two separate inequalities.
1. $x - 1 < 5x - 10 \Rightarrow 4x > 9 \Rightarrow x > 9/4$.
2. $x - 1 > -5x + 10 \Rightarrow 6x > 11 \Rightarrow x > 11/6$.
The functional intersection of these continuous zones resolves conclusively to $\mathbf{x > 9/4}$.
Alternative Algebraic Method: If both sides are guaranteed mathematically positive, square identically: $(x-1)^2 < (5x-10)^2$, avoiding absolute mechanics entirely.
3. Solving via Piecewise Breakdown
The situation becomes significantly more complex when multiple isolated absolute values are involved or variables physically exist exterior to the modulus brackets. The zeros of the individual absolute values must first be securely found to investigate the mathematical system across linear segmented cases.
EXAMPLE 3
Solve $|1 - x| = 3x + 2$.
Because $1-x$ holds positive, the modulus leaves it raw: $1 - x = 3x + 2 \Rightarrow 4x = -1 \Rightarrow x = -1/4$. Since $-1/4 < 1$, it is mathematically accepted.
Because $1-x$ holds negative, the modulus negates it cleanly: $x - 1 = 3x + 2 \Rightarrow 2x = -3 \Rightarrow x = -3/2$. Since $-3/2$ is not $> 1$, this branch is mathematically rejected.
EXAMPLE 4
Solve $|x - 1| + |x - 2| = x$.
Both brackets evaluate negatively: $-(x-1) - (x-2) = x \Rightarrow -x+1 -x+2 = x \Rightarrow 3x = 3 \Rightarrow x=1$. Since constraint explicitly requires $x < 1$, this fails and is logically rejected.
First bracket shifts positively, second holds negative: $(x-1) - (x-2) = x \Rightarrow x-1 -x+2 = x \Rightarrow \mathbf{x = 1}$. Satisfies domain constraint perfectly.
Both brackets shift positively: $(x-1) + (x-2) = x \Rightarrow 2x - 3 = x \Rightarrow \mathbf{x = 3}$. Satisfies domain constraint perfectly.
EXAMPLE 5 (Graphing Linear Modulus Sequences)
Sketch the graph of $f(x) = |x - 1| + |x - 2| - x$.
The defined graph inherently constitutes straight connected linear segments shifting purely at the identified operational zeroes $x=1$ and $x=2$. Finding four key data points plots the graph precisely:
- At $x=1$: $f(1) = 0 + |-1| - 1 = 0$
- At $x=2$: $f(2) = |1| + 0 - 2 = -1$
- Exterior left point $x=0$: $f(0) = |-1| + |-2| - 0 = 3$
- Exterior right point $x=4$: $f(4) = |3| + |2| - 4 = 1$
Connecting coordinates $(0,3), (1,0), (2,-1)$, and $(4,1)$ physically renders the continuous structured geometry of the equation.