MATHEMATICS · QUADRATIC INEQUALITIES
Solving a quadratic inequality
To solve x² − 5x + 6 ≤ 0, we need to find where the graph lies on or below the horizontal axis. The roots tell us where to begin.
f(x) ≈ -0.25
This x is in the solution.
Check the sign between the roots
x² − 5x + 6 = (x − 2)(x − 3). The roots divide the number line into three regions.
| Region | Test value | Signs | Product |
|---|---|---|---|
| x < 2 | x = 0 | (−)(−) | positive |
| 2 < x < 3 | x = 2.5 | (+)(−) | negative |
| x > 3 | x = 4 | (+)(+) | positive |
For ≤ 0, choose the negative interval and include both roots because equality is allowed. Therefore x ∈ [2, 3]. For ≥ 0, choose the outside intervals and include the roots: x ∈ (−∞, 2] ∪ [3, ∞).
What if the x² term is negative?
For −x² + 5x − 6 ≤ 0, multiply by −1. The inequality reverses: x² − 5x + 6 ≥ 0. Therefore x ∈ (−∞, 2] ∪ [3, ∞). Forgetting to reverse the sign is the central trap.
Why finding the roots is only the first step
Roots are boundary points where the expression equals zero. An inequality usually asks for intervals, not only those points.
Should the roots be included?
Square brackets include 2 and 3 because ≤ and ≥ allow equality. Strict inequalities would use parentheses.
Try it yourself
1. Solve x² − x − 6 < 0.
(x − 3)(x + 2) < 0. The upward parabola is negative between its roots, so x ∈ (−2, 3). Endpoints are excluded because the inequality is strict.
2. Solve −x² + x + 6 ≥ 0.
Multiply by −1 and reverse: x² − x − 6 ≤ 0. Factor to (x − 3)(x + 2) ≤ 0, so x ∈ [−2, 3].
Foundation → Core practice → Advanced extension
From signs to conditions
Practise sign charts in M1, then include repeated roots and poles in M2.