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?

A quadratic inequality is an inequality that can be written as ax² + bx + c > 0, < 0, ≥ 0 or ≤ 0, where a ≠ 0. Its solution is generally a range of real numbers rather than a single value.

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.

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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.

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Quadratic Inequality Formula & Tricks

Important Formulas

Quadratic roots
x = (-b ± √(b² − 4ac)) / (2a)

For ax² + bx + c = 0, the roots are obtained using the quadratic formula.

Discriminant
D = b² − 4ac

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.

Sign pattern for two roots
If α < β: for a > 0, ax² + bx + c > 0 on (-∞, α) ∪ (β, ∞) and < 0 on (α, β).

For a < 0, these positive and negative intervals are reversed. Include α and β for ≥ or ≤ inequalities.

Quick Tricks

Use the leading coefficient to set the sign pattern

After finding two distinct roots, do not test many values. If a > 0, write +, −, + from left to right. If a < 0, write −, +, −.

Example: For −x² + 4x − 3, the roots are 1 and 3. Since a < 0, the signs are negative outside [1, 3] and positive between the roots.
Factor before using the quadratic formula

If the quadratic factors easily, its roots can be found faster and the sign intervals become direct.

Example: x² − 7x + 12 = (x − 3)(x − 4), so the critical values are 3 and 4.
Check the inequality symbol at the roots

Use parentheses for strict signs > and < because the roots are excluded. Use brackets for ≥ and ≤ because the roots are included.

Example: For (x − 2)(x − 5) > 0, the answer is x < 2 or x > 5. For the same expression ≥ 0, the answer is x ≤ 2 or x ≥ 5.

Quadratic Inequality Concepts

Finding Critical Values

Critical values are the roots obtained by replacing the inequality sign with an equality sign and solving the resulting quadratic equation.

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).

Example: For 2x² − 5x − 3 < 0, 2x² − 5x − 3 = (2x + 1)(x − 3), so the critical values are −1/2 and 3.

Sign Chart for Two Distinct Real Roots

If α < β are two distinct roots, the quadratic has the same sign as a outside the roots and the opposite sign between them.

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.

Example: For 2x² − 5x − 3 < 0, the leading coefficient is positive, so the expression is negative between −1/2 and 3. Hence, −1/2 < x < 3.

Cases Based on the Discriminant

The discriminant determines whether the quadratic crosses the x-axis, touches it, or never meets it.

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.

Example: For x² + 2x + 5, D = 2² − 4(1)(5) = −16. Since a > 0, x² + 2x + 5 is positive for every real x.

Strict and Non-Strict Quadratic Inequalities

The symbols > and < exclude roots, while ≥ and ≤ include roots that make the quadratic equal to zero.

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.

Example: For (x − 1)(x + 4) ≤ 0, the solution is −4 ≤ x ≤ 1.

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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Quick Revision Notes

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.

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