Inequalities and Absolute Value
Sign Diagrams for Quadratic and Rational Inequalities
Why a quadratic inequality cannot be solved by the methods that work on a linear one, and how a sign table settles it in three steps.
What this page covers
- Build a sign diagram from the zeros of each factor
- Solve a quadratic inequality using the diagram
- Extend the method to rational inequalities with a variable denominator
- Handle the endpoints correctly, including excluded values
Why a new method is needed
For 2x - 5 > 3 the variable can be isolated. For (x - 1)(x + 2) > 0 it cannot: there is no single operation that undoes a product.
Nor is the zero-product property any help. It says a product is zero exactly when a factor is; it says nothing about when a product is positive. The relevant fact is different.
A product is positive when an even number of its factors are negative, and negative when an odd number are. Two factors: positive when they agree in sign, negative when they differ.
A factor can only change sign where it is zero. So the line splits into intervals at those points, and within each interval every factor keeps a fixed sign — and therefore so does the product. That is the whole principle.
The three steps
- Find the critical values. Set the expression to zero and solve; for a rational expression, also find where the denominator is zero.
- Draw the sign diagram. One row per factor, one column per interval. Fill in each factor's sign, then multiply down the columns for the result row.
- Read off the answer. Take the intervals whose result sign matches the inequality, then decide each endpoint separately.
Rearrange to compare against zero before starting. x2 > 3x must become x2 - 3x > 0, then x(x - 3) > 0. A sign diagram says nothing about when a product exceeds 3.
Worked example: a quadratic inequality
Solve (x - 1)(x + 2) > 0
Step 1. Zeros at x = 1 and x = -2.
Step 2. Three intervals: x < -2, -2 < x < 1, x > 1.
| Factor | x < -2 | -2 < x < 1 | x > 1 |
|---|---|---|---|
| x - 1 | − | − | + |
| x + 2 | − | + | + |
| Product | + | − | + |
Step 3. The product is positive on the outer two intervals. The inequality is strict, so the zeros themselves are excluded.
Both endpoints hollow, because the inequality is strict.
Test one value per interval: at x = -3, (-4)(-1) = 4 > 0 ✓. At x = 0, (-1)(2) = -2 < 0 ✓. At x = 2, (1)(4) = 4 > 0 ✓.
Solve x2 - 5x + 6 ≤ 0
Factor first: (x - 2)(x - 3) ≤ 0, so the critical values are 2 and 3.
| Factor | x < 2 | 2 < x < 3 | x > 3 |
|---|---|---|---|
| x - 2 | − | + | + |
| x - 3 | − | − | + |
| Product | + | − | + |
The product is negative on the middle interval. Because the inequality is ≤, the zeros are included.
Solid endpoints: the inequality admits equality.
The typed supplementary sheet illustrates this same shape with a graph: the parabola dips below the axis between its two roots and is positive outside them. The sign diagram is that picture in table form.
Rational inequalities
With a variable in the denominator, the denominator's zeros are critical values too — the expression changes sign there by becoming undefined and switching from one infinite branch to the other.
Multiplying x(2x + 5)x - 3 ≥ 0 by x - 3 requires knowing its sign. It is positive for x > 3 and negative for x < 3, so the inequality would need to be reversed on one side and not the other. The sign diagram handles both at once.
Solve x(2x + 5)x - 3 ≥ 0
Critical values: numerator zero at x = 0 and x = -52; denominator zero at x = 3.
| Factor | x < -52 | -52 < x < 0 | 0 < x < 3 | x > 3 |
|---|---|---|---|---|
| x | − | − | + | + |
| 2x + 5 | − | + | + | + |
| x - 3 | − | − | − | + |
| Result | − | + | − | + |
The result is non-negative on the second and fourth intervals. Endpoints need individual attention:
| Value | Expression there | Included? |
|---|---|---|
| x = -52 | 0 | Yes — the inequality is ≥ |
| x = 0 | 0 | Yes — same reason |
| x = 3 | Undefined | No — excluded whatever the inequality says |
Solid at the numerator's zeros, hollow at the denominator's — which is excluded regardless.
Repeated factors
A factor appearing to an even power does not change sign at its zero; it touches zero and returns. The sign diagram handles this automatically once the multiplicity is written in.
| Factor | At its zero | Sign changes? |
|---|---|---|
| (x - a) | Crosses zero | Yes |
| (x - a)2 | Touches zero | No |
| (x - a)3 | Crosses zero | Yes |
| (x - a)n, n even | Touches | No |
| (x - a)n, n odd | Crosses | Yes |
So (x - 1)2(x + 2) > 0 is positive for all x > -2 except x = 1, where the expression is zero and the strict inequality fails. The answer is (-2, 1) ∪ (1, ∞) — a single point removed from the middle.
Common mistakes
| Mistake | Correct | Why |
|---|---|---|
| Dividing by a factor | Build a sign diagram | Its sign is unknown and may reverse the inequality |
| Multiplying out a variable denominator | Include it as a row | Same reason |
| Including the denominator's zero | Always exclude it | The expression is undefined there |
| Not rearranging to compare with zero | x2 > 3x becomes x(x - 3) > 0 | The sign rule applies to a product against zero |
| Assuming the sign alternates every interval | Even multiplicities do not change sign | (x - 1)2 touches without crossing |
| Giving a single interval when two are needed | (-∞, -2) ∪ (1, ∞) | The union is part of the answer |
Frequently asked questions
Why not just divide by a factor?
Because its sign is unknown and may change. Dividing x(2x + 5) ≥ 0 by x is illegitimate: for x < 0 the division would also reverse the inequality, and for x = 0 it is undefined.
What are the critical values?
Every value at which a factor changes sign — the zeros of the numerator, and the zeros of the denominator for a rational expression. Between consecutive critical values the sign cannot change.
Is the denominator's zero ever included?
Never. It makes the expression undefined, so it is excluded regardless of whether the inequality is strict. The numerator's zeros are included when the inequality is ≤ or ≥.
Can I multiply out the denominator instead?
Only if you know its sign, which usually means splitting into cases. The sign diagram handles both cases at once and is less error-prone.
Source. Handwritten teaching notes, Week 3, page 12, and Week 4, page 2, with the typed supplementary sheet on sign diagrams.
This page is an original exposition prepared for the KEVOS® knowledge library. It restates, reorganises and verifies the mathematics of the supplied teaching notes; it is not a reproduction of them. Numerical values taken from the notes are identified as source examples and are not presented as engineering standards.