Inequalities & Modulus (AA HL)
Inequalities compare expressions rather than equate them, and the modulus (absolute value) measures distance from zero regardless of sign. This topic covers solving modulus equations and inequalities, and solving quadratic, cubic and rational inequalities using critical values and sign analysis - both graphically and algebraically.
What the syllabus says
This topic maps onto two HL-only points in the official IB Analysis & Approaches syllabus, building on the prior-knowledge requirement of solving quadratic inequalities.
| Code | Syllabus content |
|---|---|
| AHL2.15 | Solutions of \(g(x)\ge f(x)\), both graphically and analytically. Graphical or algebraic methods for simple polynomials up to degree 3. Use of technology for these and other functions. |
| AHL2.16 | The graphs of the functions \(y=|f(x)|\), \(y=f(|x|)\), \(y=\tfrac{1}{f(x)}\), \(y=f(ax+b)\), \(y=[f(x)]^2\). Solution of modulus equations and inequalities. |
Solving quadratic equations and inequalities with rational coefficients is listed as HL prior knowledge, alongside the AHL2.15 and AHL2.16 extension content above.
Key terms
Five words worth knowing cold before you touch the formulas below - each with a worked example showing exactly what it means.
What does modulus (\(|x|\)) mean?
\(|x|\) is the distance of \(x\) from zero on the number line, so it's always non-negative. Formally, \(|x|=x\) when \(x\ge0\), and \(|x|=-x\) when \(x<0\). Solving an equation or inequality with modulus signs means considering both what happens when the inside is positive and when it's negative.
e.g. \(|-7|=7\) and \(|7|=7\) - both a distance of \(7\) from zero.
How do you solve a "less than" modulus inequality like \(|x-2|<5\)?
\(|x-2|<5\) means \(x-2\) is within \(5\) of zero, so it becomes the compound inequality \(-5 e.g. \(|x-2|<5\Rightarrow -3
How do you solve a "greater than" modulus inequality like \(|x-2|>5\)?
A "greater than" modulus inequality splits into two separate branches: \(x-2>5\) or \(x-2<-5\), giving \(x>7\) or \(x<-3\) - everything outside the interval, not between two bounds. This is the exact complement of the "less than" case above.
e.g. \(|x-2|>5\Rightarrow x<-3\) or \(x>7\).
What are critical values in a sign-analysis inequality?
Critical values are the \(x\)-values where an expression equals zero or is undefined - the points where the expression's sign can change. Between consecutive critical values, a factorised expression keeps a single, constant sign, so testing one point in each interval tells you the sign of the whole interval.
e.g. For \(\dfrac{x-3}{x+1}\), the critical values are \(x=3\) (zero) and \(x=-1\) (undefined).
Why can't you cross-multiply a rational inequality by \((x-a)\) without care?
Multiplying both sides of an inequality by an expression flips the inequality sign if that expression is negative - but its sign isn't known in advance, so multiplying by \((x-a)\) directly is unsafe. Instead, move everything to one side, combine into a single fraction, and use sign analysis on the critical values.
e.g. \(\dfrac{x+1}{x-2}<0\) is solved via critical values \(-1,\,2\) and sign testing, giving \(-1
Key formulas
This topic is more about a method - sign analysis - than a list of formulas. The two rules for modulus inequalities below are the closest thing to formulas it has.
Formula reference
None of these are printed in the formula booklet - modulus inequalities and sign-analysis are examined as methods you're expected to apply, not formulas to look up.
| Rule | Used for | Booklet? |
|---|---|---|
\(|x|| Solving "less than" modulus inequalities | Not in booklet | |
| \(|x|>k \iff x<-k \text{ or } x>k\) | Solving "greater than" modulus inequalities | Not in booklet |
| \(|a|=|b| \iff a=\pm b\) | Solving modulus equations | Not in booklet |
| Critical values + sign chart | Solving quadratic, cubic and rational inequalities | Not in the formula booklet - prior knowledge |
"Less than" vs "greater than" modulus inequalities
These two cases behave in opposite ways, and mixing them up is the single most common error on this topic.
| Feature | \(|x-a|| \(|x-a|>k\) | |
|---|---|---|
| Solution shape | One connected interval | Two separate branches |
| Compound form | \(-k| \(x-a<-k\) or \(x-a>k\) | |
| Region on a number line | Between the two bounds | Outside the two bounds |
| Geometric meaning | Points within distance \(k\) of \(a\) | Points further than \(k\) from \(a\) |
Solving modulus inequalities
Every modulus inequality reduces to one of two compound-inequality shapes, once you know which case you're in.
The "less than" case
\[|x-a| Rewrite as a compound inequality and solve both parts together - the result is a single interval.
The "greater than" case
\[|x-a|>k \iff x-a<-k \text{ or } x-a>k\]
Split into two separate inequalities - the result is two branches, not one interval.
Not in the formula booklet - prior knowledgeModulus equations
\[|a|=|b| \iff a=b \text{ or } a=-b\]
Squaring both sides also works, but remember to keep both roots of the resulting equation.
Not in the formula booklet - prior knowledgeSolving inequalities by sign analysis
Quadratic, cubic and rational inequalities all use the same three-step method: find the critical values, chart the sign, then select the regions you need.
Find the critical values
\[\text{expression}=0 \text{ or undefined}\]
Factorise fully - each factor's root, and any denominator's root, is a critical value.
Not in the formula booklet - prior knowledgeBuild a sign chart
\[(-\infty,c_1),\,(c_1,c_2),\,(c_2,\infty),\dots\]
Test one value in each interval between consecutive critical values to fix the sign of the whole interval.
Not in the formula booklet - prior knowledgeRational inequalities
\[\dfrac{p(x)}{q(x)}\ \bowtie\ 0\]
Never cross-multiply by an unknown-sign expression - move everything to one side and combine into a single fraction first.
Not in the formula booklet - prior knowledgeWorked examples
Two full exam-style questions, marked exactly like the real thing. Try each one yourself before checking the worked solution.
Consider \(|3x-2|\le 7.\)
(a) Solve it.
(b) State the exact length of this interval.
Worked solution
(a) \(-7\le3x-2\le7\) M1
\(\Rightarrow-5\le3x\le9\Rightarrow-\tfrac53\le x\le3.\) A1
(b) \(3-\left(-\tfrac53\right)=\tfrac{14}{3}.\) A1
Solve \(x(x-3)^2\ge 0.\)
Worked solution
Roots \(0\) (simple) and \(3\) (double). M1
Since \((x-3)^2\ge 0\), the sign follows \(x\); the product is \(\ge 0\) when \(x\ge 0\), and also \(=0\) at \(x=3.\) A1
Solution \(x\ge 0\) (with \(x=3\) already included). A1
Common mistakes
The four slip-ups that account for most of the marks lost on this topic - worth reading before you start practising, not just after you get one wrong.
- Mixing up "between" and "outside" for modulus inequalities. \(|x-a|
- Cross-multiplying a rational inequality without care. Multiplying both sides by \((x-a)\) flips the inequality when \((x-a)\) is negative - move everything to one side and use sign analysis instead.
- Forgetting to exclude a value that makes a denominator zero. It's still a critical value for the sign chart, but it can never appear in the final solution set, even with an "or equal to" inequality.
- Losing a solution branch when squaring a modulus equation. \(|a|=|b|\) gives \(a=b\) or \(a=-b\) - squaring both sides and solving the resulting quadratic recovers both, but only if you keep every root.
Using your GDC
The guide normally indexed for this topic covers complex-number modulus and argument, not real-valued inequalities, so it doesn't apply here. The syllabus explicitly allows a graphical approach to this topic instead (AHL2.15), which every GDC model supports the same basic way.
Graphing both sides of an inequality and comparing heights avoids sign-analysis errors entirely - explicitly permitted for polynomial and modulus inequalities up to degree 3 on Paper 2.
- Enter the left-hand side as \(Y_1\) and the right-hand side as \(Y_2\) (for example \(Y_1=|x-1|\) and \(Y_2=|2x+3|\)).
- Graph both functions on the same screen and use the intersection tool to find where they cross - these are the critical values.
- Trace along the graph, or read the picture directly, to see which curve is higher between and beyond the crossing points.
- Match "higher" or "lower" back to the original inequality symbol to write the final solution set.
Tip: Sketch the bell-shaped V of each modulus graph roughly by hand first, so you know what to expect before you graph - it stops you misreading which branch is which.
See the full GDC guide for more calculator models and topics.
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Inequalities & modulus questions, marked instantly like the real exam.
Quick answers
The questions students on this topic ask most often.
How do I solve a modulus inequality like |x-2|<5?
Rewrite it as a compound inequality: \(-5
What's different about |x-2|>5?
It splits into two separate cases: \(x-2>5\) or \(x-2<-5\), giving \(x>7\) or \(x<-3\). A "greater than" modulus inequality gives two branches outside an interval, not one interval between two bounds.
How do I solve a rational inequality like (x+1)/(x-2)<0?
Find the critical values where the expression is zero or undefined (here \(x=-1\) and \(x=2\)), then test the sign of the expression in each interval those values create. Never multiply both sides by \((x-2)\) directly, since its sign is unknown.
Can I use my GDC to solve an inequality?
Yes - graph both sides as separate functions, find where they intersect, then read off from the graph which side is bigger. This is explicitly allowed for polynomial inequalities up to degree 3 on Paper 2.
Sub-topics
Inequalities & Modulus broken down into its individual skills, each with its own focused page.
Related topics
More Functions topics from the same AA HL syllabus unit, in case you want to keep going.