Modulus Inequalities (AA HL)
A modulus inequality asks how far \(x\) can sit from a fixed point while staying inside (or outside) a given distance. The two shapes \(|x-a|
26 questions on this sub-topic.
The two rules
Covered under IB syllabus reference AHL2.16. Unlike a lot of AA HL algebra, neither rule below is in the formula booklet - you need to know both by heart and be able to tell instantly which one a question is asking for.
"Less than"
\(|x| A "less than" modulus inequality always gives one interval between two bounds - write it as a double inequality and solve across it.
"Greater than"
\(|x|>k \iff x<-k \text{ or } x>k\)
A "greater than" modulus inequality always splits into two separate branches - solve each case on its own and join them with "or".
Comparing two moduli directly, like \(|a|=|b| \iff a=\pm b\), or need the full syllabus wording and formula-booklet reference table? See Inequalities & Modulus. A GDC graph is a good way to check your interval numerically - see the parent topic's GDC guidance.
Worked examples
Solve \(|x-2|<5.\)
Worked solution
Remove the modulus: \(-5 < x - 2 < 5.\) M1 A1
Add 2: \(-3 < x < 7.\) A1
Solve \(|x+1|\ge3.\)
Worked solution
\(x+1 \ge 3\) or \(x+1 \le -3.\) M1 A1
\(x \ge 2\) A1 or \(x \le -4.\) A1
Solve \(|x + 2| = |2x - 1|.\)
(a)(i) Give the value with \(x<1\).
(a)(ii) Give the value with \(x>1.\)
Worked solution
(a)(i) \(x = -\tfrac13\) A1
(a)(ii) or \(x = 3.\) A1
Solve \(|4x + 3| > 5.\)
Worked solution
Split: \(4x + 3 > 5\) or \(4x + 3 < -5.\) M1 A1
Case 1: \(4x > 2 \Rightarrow x > \tfrac12.\) A1
Case 2: \(4x < -8 \Rightarrow x < -2.\) A1 - \(x<-2\)
Common mistakes
- Mixing up "between" and "outside" for modulus inequalities. \(|x-a|
- 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.
- Squaring a modulus inequality without checking it's safe first. Squaring only preserves the inequality direction when both sides are guaranteed non-negative, as in \(|x-1|<|x+3|\) - never square across an inequality involving a plain (non-modulus) expression that could be negative.
Ready to practise properly?
26 modulus-inequality questions, marked instantly like the real exam.
Quick answers
How do you solve |x - a| < k?
Rewrite it as the double inequality \(-k < x-a < k\), then solve for \(x\) across the whole thing at once. The result is a single interval between two bounds.
How do you solve |x - a| > k?
Split into two separate cases, \(x-a>k\) or \(x-a<-k\), and solve each one independently. The result is two separate branches joined by "or", not one interval.