Quadratic Inequalities
A quadratic inequality compares a quadratic expression with zero, such as x2 − 3x − 4 < 0, and its solution is an interval or a union of intervals read from the parabola. In brief: x2 − 3x − 4 = (x − 4)(x + 1) < 0 holds between the zeros, so the solution is −1 < x < 4. For ax2 + bx + c > 0 with a > 0 and two zeros, the solution is outside the zeros.
The key idea is that the sign of a quadratic function can change only at its zeros. Therefore, you need only to find the zeros and decide on which side the parabola is above or below the x-axis. This guide shows the parabola method and the sign chart, covers the cases with no zeros and a double zero, and solves problems with a table of all cases.
How do you solve a quadratic inequality step by step?
- Move all terms to one side so that you compare with 0.
- Find the zeros of the quadratic expression, by factoring or by the formula.
- Sketch the parabola: it opens up if a > 0 and down if a < 0, and passes through the zeros.
- Read from the sketch where the parabola is above (> 0) or below (< 0) the x-axis.
- Write the answer as an interval. Include the zeros for ≥ or ≤ and exclude them for > or <.
Worked examples
Example 1: x2 − 3x − 4 < 0
The zeros: (x − 4)(x + 1) = 0, so x = 4 and x = −1. The parabola opens up, so it is below the axis between the zeros. The solution is (−1, 4). Check x = 0: −4 < 0 ✓. Check x = 5: 25 − 15 − 4 = 6, not < 0 ✓.
Example 2: x2 − 3x − 4 ≥ 0
The same zeros, but now the parabola is on or above the axis outside the zeros: (−∞, −1] ∪ [4, ∞). The zeros are included because of ≥.
Example 3: −x2 + 6x − 5 > 0
Zeros: multiply by −1 to get x2 − 6x + 5 = 0, so x = 1 and 5. The parabola of −x2 + 6x − 5 opens downward, so it is above the axis between the zeros: (1, 5). Alternatively, multiply the inequality by −1 and reverse the sign: x2 − 6x + 5 < 0, which gives (1, 5) ✓.
Example 4: 2x2 + x ≤ 6
Move: 2x2 + x − 6 ≤ 0. The discriminant is 1 + 48 = 49, so x = (−1 ± 7)/4, which gives 3/2 and −2. Factored: (2x − 3)(x + 2) ≤ 0. The solution is [−2, 3/2].
Example 5: with a quadratic formula
x2 − 2x − 1 > 0. D = 8, zeros x = 1 ± √2 ≈ 2.414 and −0.414. The solution is (−∞, 1 − √2) ∪ (1 + √2, ∞).
What if the discriminant is zero or negative?
When D = 0, the parabola touches the axis at one point. For a > 0: x2 − 6x + 9 = (x − 3)2. This is ≥ 0 for all x, > 0 for all x except 3, ≤ 0 only at x = 3 and < 0 for no x.
When D < 0, the parabola does not meet the axis. For a > 0, the expression is always positive, as for x2 + 2x + 5 = (x + 1)2 + 4. Then x2 + 2x + 5 > 0 holds for all real x, and x2 + 2x + 5 < 0 has no solution. For a < 0 and D < 0, the expression is always negative.
Solution table for all cases
| Discriminant | ax² + bx + c > 0 | ax² + bx + c ≥ 0 | ax² + bx + c < 0 | ax² + bx + c ≤ 0 |
|---|---|---|---|---|
| D > 0 | x < x₁ or x > x₂ | x ≤ x₁ or x ≥ x₂ | x₁ < x < x₂ | x₁ ≤ x ≤ x₂ |
| D = 0 | all x except x₁ | all real x | no solution | x = x₁ only |
| D < 0 | all real x | all real x | no solution | no solution |
For a < 0, multiply the inequality by −1 and reverse the sign, or use the same table with the roles of > and < exchanged.
How do you solve inequalities with fractions?
Do not multiply by an expression with an unknown sign. Move everything to one side, combine into one fraction and use a sign chart with the zeros of the numerator and the denominator. Example: (x − 1)/(x + 2) > 0. The critical points are 1 and −2. The fraction is positive when the numerator and the denominator have the same sign: x < −2 or x > 1. The solution is (−∞, −2) ∪ (1, ∞), with −2 excluded since the denominator would be zero.
Where do inequalities come from?
Inequality signs < and > were introduced by Thomas Harriot in a book published in 1631, after his death. The solution of quadratic inequalities by the sign of the graph followed from the coordinate geometry of Descartes. The systematic use of intervals and interval notation became common in the twentieth century.
Where are quadratic inequalities used today?
A ball thrown upward is above 5 m while −4.9t2 + 15t > 5. A business may need to find the production levels for which the profit −2x2 + 80x − 300 is positive: the zeros are 5 and 35, so profit is positive for 5 < x < 35. Safety margins and tolerance limits in engineering are often quadratic inequalities.
Which mistakes are common?
Writing the wrong region
For a > 0, > 0 means outside the zeros and < 0 means between them. Check with a test point.
Not reversing the sign when multiplying by a negative number
−x2 + 4 > 0 becomes x2 − 4 < 0.
Solving as an equation and stopping
The zeros are only the boundary. The solution is a set of intervals.
Wrong brackets
Use square brackets for ≤ and ≥, and round brackets for < and >.