Solve a modulus equation or inequality by removing each modulus according to whether its inside expression is positive or negative. For complicated expressions, find the critical points, split the number line into intervals, and solve separately on each interval.
The definition of modulus is
For a single modulus expression, standard results can be used directly:
| Problem | Equivalent condition |
|---|---|
| $ | f(x) |
| $ | f(x) |
| $ | f(x) |
For example, solve :
Therefore, or .
Now solve :
Adding and dividing by gives . Always include endpoints when the inequality is non-strict.
The inequality rules have a geometric meaning: modulus measures distance from zero. Thus, places within distance of zero, while places it outside that distance. This helps you choose between an “and” interval and an “or” pair of intervals. If the right-hand side of a modulus equation is negative, there are no real solutions because modulus is never negative.
For , the critical points are and . Interval work gives for , for , and for . The valid solutions are and . Substitution verifies both and rejects candidates outside their assumed intervals.
For several modulus terms, make an interval table first. In each interval, record the sign of every inside expression, replace each modulus consistently, and solve the resulting equation or inequality. Then check that each candidate lies in the interval whose case produced it; otherwise it is extraneous. For inequalities, combine only valid interval portions, including critical points when the original non-strict inequality allows equality.
A common misconception is using only for . You must also consider .
This is IB Mathematics AA HL, sub-topic AHL 2.16 Graphing modulus equations and inequalities. On Paper 1, show case splitting algebraically. On Paper 2, graphs may be used, but state intersection points or solution intervals clearly.
