What is a quadratic inequality?
A quadratic inequality has an x² term and a sign like <, ≤, > or ≥. Example: x² − x − 6 < 0.
It does not ask "which x makes it zero?". It asks "which x make it negative?" (or positive). The answer is usually a whole range of numbers, not one or two values.
Graph method: read the answer from the parabola
- Move everything to one side so the other side is 0.
- Find the roots of ax² + bx + c = 0 (factorise or use the formula).
- Sketch the parabola: opens up if a > 0, down if a < 0.
- Pick the part you need: above the axis for > 0, below for < 0.
x² − x − 6 = (x + 2)(x − 3). Roots −2 and 3. The curve opens up, so it is below the axis between the roots: x² − x − 6 < 0 ⇔ −2 < x < 3, and above outside: x² − x − 6 > 0 ⇔ x < −2 or x > 3.
Sign chart (table of signs)
Draw a number line and mark the roots. They cut it into parts. In each part, test one number and note the sign of each factor.
| x | x < −2 | −2 < x < 3 | x > 3 |
|---|---|---|---|
| x + 2 | − | + | + |
| x − 3 | − | − | + |
| product | + | − | + |
Keep the parts with the sign you want. This method also works for products of more factors.
Strict or not: <, ≤ and writing the answer
With < or > the roots are not included (open dots). With ≤ or ≥ they are included (filled dots).
x² − 4 ≥ 0 → x ≤ −2 or x ≥ 2. In interval notation: (−∞, −2] ∪ [2, ∞).
−2 < x < 3 is written (−2, 3).
Special cases: a < 0, no real roots, one double root
- a < 0: multiply both sides by −1 and flip the sign. −x² + 2x + 8 > 0 becomes x² − 2x − 8 < 0 → −2 < x < 4.
- D < 0 (no real roots): the curve never touches the axis. If a > 0 it is always positive: x² + 2x + 5 > 0 for all real x, and x² + 2x + 5 < 0 has no solution.
- D = 0: the curve touches the axis once. (x − 3)² ≤ 0 only when x = 3; (x − 3)² > 0 for every x except 3.
Parameter problems: "for all x"
"Find k so that x² + kx + 4 > 0 for every x." The parabola opens up, so it must stay above the axis and never touch it. That needs D < 0: k² − 16 < 0 → −4 < k < 4.
Rule: ax² + bx + c > 0 for all x ⇔ a > 0 and D < 0. ax² + bx + c < 0 for all x ⇔ a < 0 and D < 0.
Try it: predict, then check
In the 3D free play set p = −1, q = 4 and the sign to ≤. Before you look, write your answer. Then compare it with the answer under the graph. Now flip a to −1. Did your answer switch from "between" to "outside"?
At home: throw a ball straight up and count how long it stays above your head. That time window is the answer to a quadratic inequality.
Key formulas and definitions
- a > 0: ax² + bx + c < 0 ⇔ x₁ < x < x₂ (between the roots)
- a > 0: ax² + bx + c > 0 ⇔ x < x₁ or x > x₂ (outside the roots)
- Multiply by a negative number → flip the inequality sign
- ax² + bx + c > 0 for all x ⇔ a > 0 and b² − 4ac < 0
Worked examples
1. Solve x² − x − 6 < 0.
(x + 2)(x − 3) < 0. Roots −2 and 3. a > 0, so negative between the roots: −2 < x < 3.
2. Solve x² − 4 ≥ 0.
(x − 2)(x + 2) ≥ 0. Roots ±2. Outside the roots, roots included: x ≤ −2 or x ≥ 2.
3. Solve −x² + 2x + 8 > 0.
Multiply by −1 and flip: x² − 2x − 8 < 0 → (x − 4)(x + 2) < 0 → −2 < x < 4.
4. Solve x² + 2x + 5 > 0.
D = 4 − 20 = −16 < 0 and a > 0, so the curve is always above the axis. True for all real x.
5. Solve (x − 3)² ≤ 0.
A square is never negative. It equals 0 only at x = 3. So x = 3 is the only solution.
6. Find k so that x² + kx + 4 > 0 for all x.
Need D < 0: k² − 16 < 0 → (k − 4)(k + 4) < 0 → −4 < k < 4.
7. A ball's height is h = 20t − 5t² m. When is it above 15 m?
20t − 5t² > 15 → 5t² − 20t + 15 < 0 → t² − 4t + 3 < 0 → (t − 1)(t − 3) < 0 → 1 < t < 3 seconds.
Common mistakes
- Writing x < 3 and x < −2 from (x + 2)(x − 3) < 0, as if it were an equation. Use the graph or a sign chart.
- Forgetting to flip the sign when multiplying by a negative number.
- Writing −2 > x > 3 for the outside part. It must be two pieces: x < −2 or x > 3.
- Including the roots for a strict sign (< or >). Use open brackets ( ) then.