Quadratic Inequalities – Solving Inequalities and Interval Notation

1. Introduction In this lesson, you will learn how to solve quadratic inequalities. Unlike quadratic equations, which yield specific point solutions, quadratic inequalities have solution sets represented as intervals. You will master sign charts and interval notation.

2. Core Concept Explanation A quadratic inequality is an inequality that can be written in one of the following forms: ax^2 + bx + c > 0, \quad ax^2 + bx + c \ge 0, \quad ax^2 + bx + c < 0, \quad ax^2 + bx + c \le 0

Where a \neq 0. To solve it, we first find the critical values (roots of the corresponding equation ax^2 + bx + c = 0). These roots divide the number line into test intervals. The sign of the expression in each interval determines the solution set.

3. Step-by-Step Method To solve a quadratic inequality: 1. Rewrite the inequality so that one side is zero (standard form). 2. Find the critical values by solving the quadratic equation ax^2 + bx + c = 0. 3. Plot the critical values on a number line to divide it into three intervals. 4. Select a test point from each interval and substitute it into the expression ax^2 + bx + c to check its sign (positive or negative). 5. State the solution intervals that satisfy the original inequality. 6. Write the final answer in interval notation or set-builder notation.

4. Worked Examples * Example 1: Solve x^2 - 5x + 6 > 0. * Solution: 1. Solve x^2 - 5x + 6 = 0 \Rightarrow (x - 2)(x - 3) = 0. Critical values: x = 2 and x = 3. 2. Number line intervals: (-\infty, 2), (2, 3), and (3, \infty). 3. Test points: * For (-\infty, 2), choose x = 0: 0^2 - 5(0) + 6 = 6 > 0 (True). * For (2, 3), choose x = 2.5: 2.5^2 - 5(2.5) + 6 = -0.25 < 0 (False). * For (3, \infty), choose x = 4: 4^2 - 5(4) + 6 = 2 > 0 (True). 4. Solution: x < 2 or x > 3. 5. Interval notation: (-\infty, 2) \cup (3, \infty). * Example 2: Solve x^2 - 4 \le 0. * Solution: 1. Critical values from x^2 - 4 = 0 are x = -2 and x = 2. 2. Intervals: (-\infty, -2], [-2, 2], [2, \infty). 3. Test points: * For x = -3: (-3)^2 - 4 = 5 \not\le 0. * For x = 0: 0^2 - 4 = -4 \le 0 (True). * For x = 3: 3^2 - 4 = 5 \not\le 0. 4. Solution: -2 \le x \le 2. 5. Interval notation: [-2, 2]. * Example 3: Solve -x^2 + 3x + 4 < 0. * Solution: 1. Multiply by -1 (and flip the inequality sign): x^2 - 3x - 4 > 0. 2. Factor: (x - 4)(x + 1) = 0. Critical values: x = -1 and x = 4. 3. The solution set for > 0 will be the outer intervals. 4. Solution: x < -1 or x > 4. 5. Interval notation: (-\infty, -1) \cup (4, \infty).

5. Exam Relevance * Example JAMB Question: Solve the inequality 2x^2 - 5x - 3 \le 0. * Solution: Find critical values: 2x^2 - 5x - 3 = 0 \Rightarrow (2x + 1)(x - 3) = 0 \Rightarrow x = -0.5, 3. Since the inequality is \le 0, the solution is the closed interval between the roots. Solution: -0.5 \le x \le 3 (or [-0.5, 3]).

6. Common Mistakes * Forgetting to flip the inequality: When multiplying or dividing both sides by a negative number, you must reverse the inequality symbol. For example, -x > 2 \Rightarrow x < -2. * Writing invalid interval boundaries: Confusing brackets [, ] (inclusive, used for \le, \ge) with parentheses (, ) (exclusive, used for <, >).

7. Practice Questions 1. Solve x^2 - 9 \ge 0. * Answer: x \le -3 or x \ge 3 (or (-\infty, -3] \cup [3, \infty)). 2. Solve x^2 - 3x < 0. * Answer: 0 < x < 3 (or (0, 3)).

8. Summary / Cheat Sheet * Positive leading coefficient (a>0): ax^2 + bx + c < 0 selects the inner* interval: x_1 < x < x_2. ax^2 + bx + c > 0 selects the outer* intervals: x < x_1 or x > x_2. * Use brackets [ ] if the inequality includes equal to (\le, \ge).