Rational Inequalities
A rational inequality compares a rational expression with a number, usually zero, and it is solved with a sign chart that uses the zeros of the numerator and the denominator as boundary points. In brief: to solve (x − 1)/(x + 2) > 0, mark −2 and 1 on the number line and check the sign of the fraction on each interval. The solution is x < −2 or x > 1. Never multiply by an expression whose sign you do not know.
The sign of a fraction can change only at a zero of the numerator or at a zero of the denominator. At the zero of the denominator the fraction is undefined, so this value is always excluded. This guide shows the method step by step, how to handle inequalities that need rewriting, repeated factors and non-strict inequalities, and gives a sign table and worked examples.
How do you solve a rational inequality step by step?
- Move all terms to one side so that the other side is 0.
- Combine into a single fraction and factor the numerator and the denominator.
- Find the critical points: zeros of the numerator and zeros of the denominator.
- Mark them on a number line; they split it into intervals.
- Test one point in each interval, or use the signs of the factors, to find the sign of the fraction.
- Choose the intervals with the required sign. Zeros of the numerator are included for ≥ or ≤, but zeros of the denominator are never included.
Worked examples
Example 1: (x − 1)/(x + 2) > 0
Critical points: x = 1 (numerator) and x = −2 (denominator). Intervals: (−∞, −2), (−2, 1) and (1, ∞). Test x = −3: (−4)/(−1) = 4 > 0. Test x = 0: (−1)/2 < 0. Test x = 2: 1/4 > 0. The solution is (−∞, −2) ∪ (1, ∞).
Example 2: (x − 1)/(x + 2) ≤ 0
The fraction is negative on (−2, 1), and zero at x = 1. The point x = −2 is excluded. The solution is (−2, 1].
Example 3: with factored quadratics
(x2 − 4)/(x − 3) ≥ 0. Factor: (x − 2)(x + 2)/(x − 3). Critical points: −2, 2 (included) and 3 (excluded). Signs: x < −2: (−)(−)/(−) = −. −2 < x < 2: (−)(+)/(−) = +. 2 < x < 3: (+)(+)/(−) = −. x > 3: +. The solution is [−2, 2] ∪ (3, ∞).
Example 4: a constant on the right side
(2x + 1)/(x − 3) < 1. Move 1 across and combine: (2x + 1)/(x − 3) − 1 = (2x + 1 − x + 3)/(x − 3) = (x + 4)/(x − 3) < 0. Critical points: −4 and 3. The fraction is negative between them: (−4, 3). Check with x = 0: 1/(−3) = −0.33 < 1 ✓. Check with x = 4: 9/1 = 9, not < 1.
How do you rewrite an inequality with a fraction on both sides?
Do not cross-multiply unless you know the signs of the denominators. If you multiply by an expression that is negative for some x, the inequality reverses for those x only, and you would have to split into cases. The safe way is to subtract and combine into one fraction.
Example 5: 2/(x + 1) ≥ 3/(x − 2)
2/(x + 1) − 3/(x − 2) ≥ 0. Combine: [2(x − 2) − 3(x + 1)]/((x + 1)(x − 2)) = (−x − 7)/((x + 1)(x − 2)) ≥ 0. Multiply by −1 and reverse the sign: (x + 7)/((x + 1)(x − 2)) ≤ 0. Critical points: −7 (included), −1 and 2 (excluded). Signs of (x + 7)/((x + 1)(x − 2)): x < −7: (−)/((−)(−)) = −. −7 < x < −1: (+)/((−)(−)) = +. −1 < x < 2: (+)/((+)(−)) = −. x > 2: +. We need ≤ 0: (−∞, −7] ∪ (−1, 2). Check x = −8: 2/(−7) = −0.29 and 3/(−10) = −0.3, and −0.29 ≥ −0.3 ✓.
What about repeated factors?
A factor raised to an even power does not change the sign when you pass through its zero. Example: (x − 1)2/(x + 3) > 0. Critical points: 1 (double zero of the numerator) and −3 (excluded). For x > −3 the denominator is positive and the numerator is positive except at x = 1, where it is zero. So the solution is (−3, 1) ∪ (1, ∞). For x < −3 the fraction is negative. A zero of an odd power changes the sign.
Sign chart for (x − 1)/(x + 2)
| Interval | x − 1 | x + 2 | Fraction |
|---|---|---|---|
| x < −2 | − | − | + |
| x = −2 | − | 0 | undefined |
| −2 < x < 1 | − | + | − |
| x = 1 | 0 | + | 0 |
| x > 1 | + | + | + |
Where do these methods come from?
The study of inequalities began to be systematic in the seventeenth century, with the introduction of the signs < and > by Thomas Harriot. The sign chart is an application of the intermediate value property of continuous functions: a function that is continuous on an interval and has no zero there has a constant sign. The rational function is continuous on every interval that does not contain a zero of its denominator.
Where are they used today?
Rational inequalities determine ranges where an average is below a target. For the average cost C(n) = (500 + 3n)/n, the condition C(n) < 5 gives (500 + 3n)/n − 5 < 0, so (500 − 2n)/n < 0. For n > 0, this holds when n > 250. Similar inequalities appear in concentration problems and in electronics, where conditions on a parallel resistance lead to a sign analysis.
Which mistakes are common?
Multiplying by the denominator
If the denominator can be negative, multiplying by it changes the direction of the inequality for some x. Use a sign chart instead.
Including a zero of the denominator
The fraction is undefined there. Always use a round bracket.
Forgetting that a double zero does not change the sign
With (x − 1)2, the sign is the same on both sides of x = 1.
Not moving all terms to one side first
Compare the fraction with 0. Only then do the signs of the factors tell the answer.