To solve a quadratic inequality, rearrange so that one side is zero, find the roots, sketch the graph, and read off where it is above or below the axis. The answer is an interval between the roots or two regions outside them.
This lesson is part of quadratic functions. It uses the turning-point and root skills from completing the square.
What is the method?
Follow four steps in order.
- Rearrange so that one side is zero.
- Factorise and find the roots.
- Sketch the graph, marking the roots.
- Read the region that matches the inequality sign.
The sketch is what stops the guesswork. You do not need an accurate drawing, only the shape and the two crossing points.
Worked example: greater than zero
Here is an original example. Solve x² − x − 6 > 0.
- Already zero on the right.
- Factorise: (x − 3)(x + 2) > 0, so the roots are x = 3 and x = −2.
- The x² term is positive, so the graph opens upward and crosses the axis at −2 and 3.
- We want above the axis (greater than zero), which is outside the roots.
So x < −2 or x > 3.
Test a value in the middle, x = 0: 0 − 0 − 6 = −6, which is not above zero. A value outside, x = 4: 16 − 4 − 6 = 6, which is above zero. The test agrees.
Worked example: less than or equal to zero
Solve x² − 5x + 4 ≤ 0.
The factors are (x − 1)(x − 4) ≤ 0, with roots 1 and 4. The graph opens upward, and we want on or below the axis, which is between the roots.
So 1 ≤ x ≤ 4. Include the roots because the sign is ≤.
When the inequality is not already in the right form
Solve x(x + 1) > 6.
Wrong route. Divide by x, or set x > 6 or x + 1 > 6. That treats a product as if each factor had to exceed 6, which is false.
Right route. Expand and move everything to one side: x² + x − 6 > 0. Factorise: (x + 3)(x − 2) > 0. The roots are −3 and 2, and the graph opens upward.
We want above the axis, so x < −3 or x > 2.
Check x = 3: 3 × 4 = 12, which is above 6. Check x = 0: 0, which is not above 6. The answer holds.
The mistake that costs marks
The common slip is to write the answer to the first example as −2 < x > 3. It looks like a compact form of “outside the roots”, but the statement cannot be read sensibly.
| Step | Wrong | Right |
|---|---|---|
| Two regions outside the roots | −2 < x > 3 | x < −2 or x > 3 |
| One region between the roots | x < −2 and x > 3 | −2 < x < 3 |
| Meaning | Says x > 3 only | Two separate regions |
For “between”, a single combined statement is fine. For “outside”, two statements joined by “or” are needed.
Check yourself
Solve 2x² − x − 3 ≤ 0.
Answer
Factorise: 2x² − x − 3 = (2x − 3)(x + 1). The roots are x = 3/2 and x = −1.
The graph opens upward, and ≤ 0 means on or below the axis, which is between the roots.
−1 ≤ x ≤ 3/2.
Check x = 0: −3 ≤ 0, true. Check x = 2: 8 − 2 − 3 = 3, which is not ≤ 0, so outside the interval. Consistent.
What to study next
All four skills in this chapter now work together. Try the mixed quadratic functions practice set, and use the quadratic graph and roots explorer to see the regions appear as you change the numbers.
If you want a teacher to have you sketch before you write, see online one-to-one Additional Mathematics tuition.