Maths › Further Pure 1 › Inequalities and inequations
Inequalities and inequations
Never multiply an inequality by something whose sign you do not know. Move everything to one side, factorise, and let a sign diagram or a sketch read off the answer.
Builds on Simultaneous equations and inequalities and Functions, inverses and the modulus.
IN THIS TOPIC
- Solve rational inequalities by collecting on one side and using a sign diagram.
- Handle modulus inequalities by squaring or by splitting into cases.
- State solution sets correctly, excluding the values that break a denominator.
- Use a sketch as a check on an algebraic answer.
COMMON MISCONCEPTION
To solve x/(x − 2) < 3 you simply multiply both sides by x − 2 and solve the linear inequality.
Why the obvious move is fatal
Multiplying an inequality by a negative number reverses it, and x − 2 is negative for every x below 2. Multiplying straight through by x − 2 assumes a sign it has no right to, and it silently loses the whole interval x < 2. Take the safe route instead. Subtract, combine over a common denominator, factorise, and read the signs off.
WORKED EXAMPLE
The right way round
Solve x/(x − 2) < 3.
Subtract: x/(x − 2) − 3 < 0, so (x − 3(x − 2))/(x − 2) < 0, that is (6 − 2x)/(x − 2) < 0.
The critical values are x = 2 and x = 3. Testing each interval gives negative for x < 2, positive between them, negative for x > 3.
Solution: x < 2 or x > 3. The naive multiplication would have returned only x > 3, throwing away half the answer.
Critical values come from zeros of the numerator and of the denominator alike, and the two behave differently. A numerator zero may be included when the inequality is weak. A denominator zero is always excluded, since the expression has no value there.
Modulus and the sketching route
For an inequality between a modulus and something non-negative, squaring is legitimate and usually quickest, because both sides are then non-negative and the direction is preserved. Otherwise split into cases on the sign inside the modulus. Either way, sketch both sides afterwards and read off the intervals where one curve sits above the other, since that catches a lost interval in seconds.
WORKED EXAMPLE
A modulus against a line
Solve |x² − 1| > 2(x + 1).
Where x² ≥ 1 the modulus opens directly: x² − 1 > 2x + 2 gives x² − 2x − 3 > 0, that is (x − 3)(x + 1) > 0, so x > 3 or x < −1, both of which lie inside this case.
Where x² < 1 it opens with a sign change: 1 − x² > 2x + 2 gives x² + 2x + 1 < 0, which is (x + 1)² < 0 and therefore impossible.
Solution: x < −1 or x > 3. The sketch agrees. The modulus graph dips to the axis at x = ±1 and rises above the line only outside the two crossings.
INDEPENDENT PRACTICE
Two fractions at once
Solve 1/(x − 1) > x/(x + 2).
Show the working
Subtract and combine: (x + 2 − x(x − 1))/((x − 1)(x + 2)) > 0, which is (2 + 2x − x²)/((x − 1)(x + 2)) > 0.
The numerator vanishes at x = 1 ± √3 and the denominator at x = 1 and x = −2, giving four critical values.
Testing the five intervals leaves −2 < x < 1 − √3 or 1 < x < 1 + √3. Both denominator zeros stay excluded throughout.
ASSESSMENT FOCUS
- Never multiply by an expression of unknown sign; subtract to one side instead.
- List every critical value, from numerator and denominator, before testing any interval.
- Exclude denominator zeros from the solution set even when the inequality is weak.
- A quick sketch of both sides is worth the thirty seconds, and it catches lost intervals immediately.
CHECK YOURSELF
Solve (x + 1)/(x − 3) ≥ 0.
Show a hint
Critical values −1 and 3; test the three intervals and mind which endpoint can be included.
Show the answer
The expression is positive for x < −1 and for x > 3, and zero at x = −1. So x ≤ −1 or x > 3. The numerator zero is included by the weak inequality, and the denominator zero never is.
Collect on one side, factorise, and read a sign diagram; multiplying by an unknown sign loses intervals.
Square a modulus inequality when both sides are non-negative, otherwise split into cases and sketch to check.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their worked answers on the inequalities and inequations questions page.
CHECK YOUR PROGRESS
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- Solve rational inequalities by collecting on one side and using a sign diagram.
- Handle modulus inequalities by squaring or by splitting into cases.
- State solution sets correctly, excluding the values that break a denominator.
- Use a sketch as a check on an algebraic answer.
Open the full revision checklist to see every objective in the course in one place.