The Modulus Function

$|x|$ is the distance of $x$ from zero, so it discards the sign. That single idea explains why every modulus equation splits into two cases and why every modulus inequality has two shapes.

Definition and graph

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

So $|7| = 7$ and $|-7| = 7$. Equivalently,

$$|x| = \sqrt{x^2}$$

which is often the most useful form, because squaring is how you remove a modulus safely.

The graph of $y = |x|$ is a V with its vertex at the origin: the line $y = x$ for $x \ge 0$, and its reflection $y = -x$ for $x < 0$.

Graphs of modulus functions

$y = |f(x)|$ — sketch $y = f(x)$, then reflect every part below the $x$-axis up above it. Nothing above the axis moves.

So $y = |x - 2|$ is the line $y = x - 2$ with the part left of $x = 2$ folded upwards, giving a V with vertex $(2, 0)$.

And $y = |x^2 - 4|$ is the parabola with the section between $x = -2$ and $x = 2$ flipped above the axis, producing a W-like shape with two sharp corners at $(\pm2, 0)$.

$y = f(|x|)$ is different: keep the part of the graph for $x \ge 0$ and reflect it in the $y$-axis, discarding whatever was there before. The result is always symmetric about the $y$-axis.

Solving modulus equations

$|A| = k$ (with $k \ge 0$) means $A = k$ or $A = -k$. Two cases, both to be solved.

Solve $|2x - 6| = 4$.

$$2x - 6 = 4 \quad \Rightarrow \quad x = 5$$ $$2x - 6 = -4 \quad \Rightarrow \quad x = 1$$

Check both: $|2(5) - 6| = |4| = 4$ ✓ and $|2(1) - 6| = |-4| = 4$ ✓

Equations with two moduli

When both sides carry a modulus, squaring both sides is cleanest: it removes both at once, and it is valid because $|A|^2 = A^2$.

Solve $|x - 3| = |2x + 1|$.

$$(x-3)^2 = (2x+1)^2$$ $$x^2 - 6x + 9 = 4x^2 + 4x + 1$$ $$0 = 3x^2 + 10x - 8$$ $$(3x - 2)(x + 4) = 0$$ $$x = \tfrac23 \quad \text{or} \quad x = -4$$

Check $x = \frac23$: $\left|\frac23 - 3\right| = \frac73$ and $\left|\frac43 + 1\right| = \frac73$ ✓

Check $x = -4$: $|-7| = 7$ and $|-8+1| = 7$ ✓

Modulus inequalities

Two shapes, and they behave quite differently.

$|A| < k$ means $-k < A < k$ — a single interval, "within $k$ of zero".

Solve $|x - 2| < 5$.

$$-5 < x - 2 < 5$$ $$-3 < x < 7$$

Read it as: $x$ is less than 5 away from 2.

$|A| > k$ means $A > k$ or $A < -k$ — two separate intervals, "further than $k$ from zero".

Solve $|x + 1| > 3$.

$$x + 1 > 3 \quad \Rightarrow \quad x > 2$$ $$x + 1 < -3 \quad \Rightarrow \quad x < -4$$

The solution is $x < -4$ or $x > 2$ — it cannot be written as one inequality, because the values in between are excluded.

Solve $|2x - 1| \le 7$.

$$-7 \le 2x - 1 \le 7$$ $$-6 \le 2x \le 8$$ $$-3 \le x \le 4$$

The triangle inequality

$$|a + b| \le |a| + |b|$$

The two sides are equal when $a$ and $b$ have the same sign, and the left is strictly smaller when they have opposite signs, because some cancellation occurs first.

$$|3 + 5| = 8 = |3| + |5| \qquad |3 + (-5)| = 2 < |3| + |-5| = 8$$

A companion result:

$$\big||a| - |b|\big| \le |a - b|$$