Quadratic Inequality: Formula, Rules and Solved Examples
Quadratic Inequality involves comparing a quadratic expression with zero or another expression using signs such as >, <, ≥ or ≤. Its solution is found by calculating the roots, placing them on a number line and checking the sign in each interval. This page covers the standard formula, sign rules, special cases and solved aptitude examples.
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What Is a Quadratic Inequality?
First find the critical values by solving ax² + bx + c = 0. If the two roots are α and β, with α < β, the sign of the quadratic changes at these points. When a > 0, the expression is positive outside the roots and negative between them. When a < 0, the signs are reversed. For example, x² − 5x + 6 ≥ 0 becomes (x − 2)(x − 3) ≥ 0, so x ≤ 2 or x ≥ 3.
Quadratic Inequality Video Lessons
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Quadratic Equations: Comparing Roots
Learn how to compare the roots of quadratic equations and determine which root is larger using algebraic relationships and sign-based reasoning.
Quadratic Inequality Formula & Tricks
Important Formulas
For ax² + bx + c = 0, the roots are obtained using the quadratic formula.
D determines the number and type of real roots: D > 0 gives two distinct real roots, D = 0 gives one repeated real root, and D < 0 gives no real roots.
For a < 0, these positive and negative intervals are reversed. Include α and β for ≥ or ≤ inequalities.
Quick Tricks
After finding two distinct roots, do not test many values. If a > 0, write +, −, + from left to right. If a < 0, write −, +, −.
If the quadratic factors easily, its roots can be found faster and the sign intervals become direct.
Use parentheses for strict signs > and < because the roots are excluded. Use brackets for ≥ and ≤ because the roots are included.
Quadratic Inequality Concepts
Finding Critical Values
For ax² + bx + c ≥ 0, first solve ax² + bx + c = 0. The roots divide the number line into intervals where the expression has a constant sign. The roots may be found by factorisation or by x = (-b ± √(b² − 4ac))/(2a).
Sign Chart for Two Distinct Real Roots
For a > 0, the sign pattern is positive, negative, positive across the intervals (−∞, α), (α, β), and (β, ∞). For a < 0, it is negative, positive, negative. Select the intervals matching the required inequality and include endpoints only for non-strict inequalities.
Cases Based on the Discriminant
If D > 0, there are two distinct real roots and three sign intervals. If D = 0, the quadratic has a repeated root and does not change sign at that root. If D < 0, the quadratic has no real roots and always has the sign of a for every real x.
Strict and Non-Strict Quadratic Inequalities
For x² − 9 > 0, the roots −3 and 3 are excluded, giving x < −3 or x > 3. For x² − 9 ≥ 0, the same intervals include the roots, giving x ≤ −3 or x ≥ 3. A zero value is never included in a strict inequality.
Quadratic Inequality Data Table
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Sign Rules for a Quadratic with Two Distinct Roots
Let α < β be the roots of ax² + bx + c.
| Condition | Outside the roots | Between the roots |
|---|---|---|
| a > 0 | Positive | Negative |
| a < 0 | Negative | Positive |
Note: At α and β, the quadratic equals zero. Include these endpoints for ≥ or ≤, but exclude them for > or <.
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Quadratic Inequality Revision Points
Use these rules to solve a quadratic inequality in interval form.
- Rewrite the inequality so that one side is zero.
- Solve the corresponding quadratic equation to obtain critical values.
- For two roots α < β, use the sign of the leading coefficient to set the sign pattern.
- If a > 0, the quadratic is positive outside the roots and negative between them.
- If a < 0, the quadratic is negative outside the roots and positive between them.
- Include roots for ≥ and ≤; exclude them for > and <.
- If D < 0, the quadratic always has the sign of a.
- If D = 0, the expression touches zero but does not change sign.
Quadratic Inequality FAQs
What is the formula for solving a quadratic inequality?
There is no separate root formula. Find the roots of the corresponding equation using x = (-b ± √(b² − 4ac))/(2a), then use the sign intervals determined by the roots and the leading coefficient.
How do you solve x² − 5x + 6 > 0?
Factor it as (x − 2)(x − 3) > 0. Since the leading coefficient is positive, the expression is positive outside the roots, so x < 2 or x > 3.
How does the answer change for x² − 5x + 6 ≥ 0?
The roots are 2 and 3, and the non-strict sign includes them. Therefore, x ≤ 2 or x ≥ 3.
What is the solution when the discriminant is negative?
The quadratic has no real roots and keeps the sign of its leading coefficient. For example, x² + 1 > 0 for every real x because D = −4 and a > 0.
What is the solution of −x² + 4x − 3 ≥ 0?
Factor it as −(x − 1)(x − 3). Since the leading coefficient is negative, the expression is non-negative between the roots, including them. Thus, 1 ≤ x ≤ 3.
Why are roots included in a quadratic inequality with ≥ or ≤?
At a root, the quadratic equals zero. Since ≥ and ≤ allow zero, the root is included. The symbols > and < do not allow zero, so their roots are excluded.
