Algorithmic Solution of Quadratic Inequalities in One Real Variable
Solving quadratic inequalities in one variable is the culmination of functional sign analysis. A quadratic inequality is an algebraic statement of the form ax² + bx + c > 0, ≥ 0, < 0, or ≤ 0 (with a ≠ 0). Unlike linear inequalities, which can be solved through straightforward algebraic isolation, quadratic inequalities require investigating the sign distribution across intervals bounded by the real zeros of the corresponding quadratic equation. The solution set is typically expressed as an open interval (x₁, x₂), a closed interval [x₁, x₂], the complement of an interval ℝ \ [x₁, x₂], a singleton set, or the empty set ∅. In this lesson, Secondary 1 scholars learn a foolproof four-step algorithm to solve any quadratic inequality with absolute precision.
1. The Standard 4-Step Algorithmic Protocol
- Standardize the Inequality: Rearrange all terms to the left-hand side so that the right-hand side is zero: ax² + bx + c ⋚ 0.
- Find the Zeros: Set the associated quadratic equation ax² + bx + c = 0 and solve for roots using factoring or the general formula.
- Sketch the Sign on the Number Line:
- If Δ > 0 with roots x₁ < x₂: Sign is opposite to a inside (x₁, x₂), and same as a outside [x₁, x₂].
- If Δ = 0 with root x₀: Sign is same as a everywhere except at x₀ where it equals 0.
- If Δ < 0: Sign is same as a everywhere on ℝ.
- Select the Solution Interval: Read off the matching interval from the number line that satisfies the inequality symbol (> 0, ≥ 0, < 0, ≤ 0).
2. Comprehensive Summary Matrix of Solution Sets
Assuming a > 0 (parabola opens upward) with two distinct real roots x₁ < x₂:
| Inequality Form | Required Sign | Solution Set (Interval Notation) |
|---|---|---|
| ax² + bx + c < 0 | Strictly Negative | (x₁, x₂) (Open interval between roots) |
| ax² + bx + c ≤ 0 | Negative or Zero | [x₁, x₂] (Closed interval including roots) |
| ax² + bx + c > 0 | Strictly Positive | ℝ \ [x₁, x₂] = (-∞, x₁) ∪ (x₂, ∞) |
| ax² + bx + c ≥ 0 | Positive or Zero | ℝ \ (x₁, x₂) = (-∞, x₁] ∪ [x₂, ∞) |
Inequality Workshop: Special Cases
Case Study: Solving x² - 4x + 4 ≤ 0
Determine the complete solution set in ℝ.
• Equation: x² - 4x + 4 = 0 ⟹ (x - 2)² = 0 ⟹ Single root x = 2 (Δ = 0).
• For any x ∈ ℝ, (x - 2)² ≥ 0.
• The inequality demands (x - 2)² ≤ 0. The only value satisfying this is where it equals zero.
• Solution Set = {2} (a single point!).
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