IB Maths AA HL Topic 2 — Functions Paper 1 & 2 HL only ~8 min read

Modulus Equations & Inequalities

A modulus equation hides two cases — one where the inside is positive, one where it’s negative. The standard move is to split into two equations and solve each, then check both answers in the original. Inequalities follow naturally: solve the equation first to find boundary points, then test intervals or sketch. Watch out for extraneous solutions — they’re the headline trap on this topic.

📘 What you need to know

For |f(x)| = g(x): split into two cases — f(x) = g(x) and f(x) = −g(x) — solve each, then check.
For |f(x)| = |g(x)|: split similarly. Both sides are non-negative, so solutions almost always work — but still verify.
Always check solutions in the original equation. Extraneous roots arise when the splitting step is reversed by the modulus.
For |f(x)| < k (positive k): equivalent to −k < f(x) < k.
For |f(x)| > k (positive k): equivalent to f(x) > k or f(x) < −k.
For |f(x)| < |g(x)|: square both sides — both are non-negative, so squaring preserves the direction.
Sketching always helps. Draw y = |f(x)| and the other side; intersections are equation solutions, regions are inequality solutions.

Solving modulus equations

Standard splitting rule |f(x)| = g(x) ⟺ f(x) = g(x) or f(x) = −g(x)

The split captures both possibilities for what the inside of the modulus could have been. Once you solve each linear (or polynomial) equation, you must check the answers in the original — the splitting can introduce values where the right-hand side is negative, which the modulus on the left can never equal.

|f| = |g|

both sides non-negative

extraneous solutions are rare; checking is still required

|f| = g

g(x) might be negative

extraneous solutions are common — always verify

Why do extraneous solutions appear? The equation f(x) = −g(x) might have a perfectly valid algebraic solution — but if that x makes g(x) negative, then |f(x)| (always ≥ 0) can’t equal g(x) (now negative). The split equation lied; the original equation didn’t have that solution.

Solving modulus inequalities

Two clean rules cover the most common cases when one side is a positive constant k:

Inequalities with constants |f(x)| < k ⟺ −k < f(x) < k
|f(x)| > k ⟺ f(x) > k or f(x) < −k

For more complex cases, solve the corresponding equation first to find the boundary points, then either test intervals or sketch both sides on the same axes.

🧭 Recipe — modulus inequality

Replace the inequality with = and solve for the boundary points (using the splitting method above).
Check each boundary in the original equation — discard any extraneous roots.
Sketch both sides of the inequality on the same axes (or test values in each interval).
Read off the regions: above-or-below depending on the inequality direction.
Match the endpoint type: ≤/≥ uses inclusive boundaries; </> uses strict.

When to square both sides

For |f(x)| ≷ |g(x)| (both sides modulus), squaring works cleanly because both sides are non-negative — squaring preserves the inequality direction:

Squaring shortcut |f(x)| < |g(x)| ⟺ f(x)² < g(x)²

Why this works: squaring is monotonic on non-negative numbers — if a, b ≥ 0 and a < b, then a² < b². After squaring, the modulus signs vanish and you’re left with a polynomial inequality you can solve directly.

Don’t square when only one side has a modulus and the other could be negative — squaring then can introduce extraneous solutions just like splitting can. Save squaring for the |f| ≷ |g| case where both sides are guaranteed non-negative.

Worked examples

WE 1

Basic modulus equation

Solve |3x − 5| = 7.

Step 1: Split into two cases 3x − 5 = 7 or 3x − 5 = −7 Step 2: Solve each 3x = 12 → x = 4 3x = −2 → x = −2/3 Step 3: Check both |3(4) − 5| = |7| = 7 ✓ |3(−2/3) − 5| = |−7| = 7 ✓ x = 4 or x = −2/3 when the right-hand side is a positive constant, both split equations almost always give valid answers

WE 2

Modulus equation with an extraneous solution

Solve |2x + 1| = 3x − 4.

Step 1: Split into two cases 2x + 1 = 3x − 4 or 2x + 1 = −(3x − 4) Step 2: Solve each first: 1 + 4 = 3x − 2x → x = 5 second: 2x + 1 = −3x + 4 → 5x = 3 → x = 3/5 Step 3: Check both in original x = 5: |11| = 11; 3(5) − 4 = 11 ✓ x = 3/5: |11/5| = 11/5; 3(3/5) − 4 = −11/5 ✗ x = 5 only x = 3/5 makes the right-hand side negative — but |2x + 1| can never equal a negative value

WE 3

Modulus equals modulus

Solve |2x − 3| = |x + 4|.

Step 1: Split into two cases 2x − 3 = x + 4 or 2x − 3 = −(x + 4) Step 2: Solve each first: x = 7 second: 2x − 3 = −x − 4 → 3x = −1 → x = −1/3 Step 3: Check both in original x = 7: |11| = 11; |11| = 11 ✓ x = −1/3: |−11/3| = 11/3; |11/3| = 11/3 ✓ x = 7 or x = −1/3 |f| = |g| almost always gives both solutions valid because both sides are non-negative

WE 4

Modulus inequality less than a constant

Solve |3x − 2| < 8.

Step 1: Apply |f(x)| < k ⟺ −k < f(x) < k −8 < 3x − 2 < 8 Step 2: Add 2 to all three parts −6 < 3x < 10 Step 3: Divide by 3 −2 < x < 10/3 −2 < x < 10/3 strict inequality preserves strict endpoints — open interval

WE 5

Modulus inequality greater than a constant

Solve |x − 5| ≥ 3.

Step 1: Apply |f(x)| ≥ k ⟺ f(x) ≥ k or f(x) ≤ −k x − 5 ≥ 3 or x − 5 ≤ −3 Step 2: Solve each x ≥ 8 or x ≤ 2 x ≤ 2 or x ≥ 8 ≥ uses inclusive endpoints; the solution is “outside” the band of values where |x − 5| < 3

WE 6

Modulus less than modulus — square both sides

Solve |x − 2| < |x + 4|.

Step 1: Square both sides (valid since both ≥ 0) (x − 2)² < (x + 4)² Step 2: Expand both sides x² − 4x + 4 < x² + 8x + 16 Step 3: Simplify −4x + 4 < 8x + 16 −12x < 12 x > −1 (dividing by −12 flips the inequality) x > −1 geometric reading: |x − 2| is distance from x to 2, |x + 4| is distance from x to −4. The inequality says x is closer to 2 than to −4 — i.e., right of the midpoint −1 ✓

💡 Top tips

Always check both split solutions in the original equation. Extraneous roots are the most common reason students lose marks.
For |f| = k (constant): split is f(x) = ±k. Both solutions usually valid when k > 0.
For |f| = g(x) (variable): split is f(x) = ±g(x). Always check — extraneous solutions appear when g(x) is negative at the candidate x.
For |f| ≷ |g|: square both sides. Reduces the modulus problem to a polynomial inequality.
Sketch when in doubt. Two graphs on the same axes — boundary points are intersections, solutions are regions where one is above/below the other.
Watch the inequality direction when you divide by a negative number — flip it.
Use the geometric interpretation: |x − a| is the distance from x to a. Often gives the shortest path to a sanity check.

⚠ Common mistakes

Skipping the verification step after splitting an equation. Extraneous solutions slip through silently.
Squaring an inequality where one side might be negative. Squaring is only safe when both sides are non-negative — i.e., for |f| ≷ |g|.
Dropping the modulus signs without splitting. |f(x)| ≠ f(x) unless f(x) ≥ 0.
Mixing up the direction of |f| > k: it’s an “or” of two regions, not a single interval.
Forgetting to flip the inequality when dividing by a negative number — common when simplifying after squaring.
Confusing strict and inclusive endpoints. Match the inequality type ≤ vs < in the answer.
Solving only one of the two split equations — half a solution is no solution.

Modulus equations and inequalities close out the modulus topic. The next note, Reciprocal Transformations, switches focus: how the graph of y = 1/f(x) relates to the original y = f(x). Roots become asymptotes; asymptotes become roots; maxes and mins flip — a different kind of transformation, but with its own clean rules.

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