Quadratic Formula Explained – Derivation, Application, and Step-by-Step Examples

1. Introduction In this lesson, you will learn about the quadratic formula, how it is derived from standard algebraic principles, and how to apply it systematically to find the roots of any quadratic equation. This formula is a universal method that works even when factorization is difficult or impossible.

2. Core Concept Explanation For any quadratic equation in the standard form: ax^2 + bx + c = 0

Where a \neq 0, the solutions (or roots) are given by the quadratic formula: x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

The term under the square root, b^2 - 4ac, is called the discriminant (\Delta). It determines the nature of the roots: * If b^2 - 4ac > 0, there are two distinct real roots. * If b^2 - 4ac = 0, there is one repeated real root. * If b^2 - 4ac < 0, there are two complex (imaginary) roots.

3. Step-by-Step Method To solve ax^2 + bx + c = 0 using the quadratic formula: 1. Rearrange the equation into standard form: ax^2 + bx + c = 0. 2. Identify the coefficients a, b, and c. Be extremely careful with negative signs. 3. Compute the discriminant \Delta = b^2 - 4ac to see what kind of roots to expect. 4. Substitute a, b, and the computed discriminant into the quadratic formula. 5. Simplify the numerator and denominator to get the two roots.

4. Worked Examples * Example 1: Solve x^2 - 5x + 6 = 0. * Solution: 1. Coefficients: a = 1, b = -5, c = 6. 2. Discriminant: b^2 - 4ac = (-5)^2 - 4(1)(6) = 25 - 24 = 1. 3. Formula: x = \frac{-(-5) \pm \sqrt{1}}{2(1)} = \frac{5 \pm 1}{2}. 4. Roots: x_1 = \frac{5 + 1}{2} = 3 and x_2 = \frac{5 - 1}{2} = 2. * Example 2: Solve 2x^2 + 5x - 3 = 0. * Solution: 1. Coefficients: a = 2, b = 5, c = -3. 2. Discriminant: b^2 - 4ac = 5^2 - 4(2)(-3) = 25 + 24 = 49. 3. Formula: x = \frac{-5 \pm \sqrt{49}}{2(2)} = \frac{-5 \pm 7}{4}. 4. Roots: x_1 = \frac{-5 + 7}{4} = 0.5 and x_2 = \frac{-5 - 7}{4} = -3. * Example 3: Solve x^2 - 4x + 1 = 0. Leave your answer in surd form. * Solution: 1. Coefficients: a = 1, b = -4, c = 1. 2. Discriminant: b^2 - 4ac = (-4)^2 - 4(1)(1) = 16 - 4 = 12. 3. Formula: x = \frac{4 \pm \sqrt{12}}{2} = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3}. 4. Roots: x_1 = 2 + \sqrt{3} and x_2 = 2 - \sqrt{3}.

5. Exam Relevance * Example JAMB Question: Solve for x in 3x^2 - 7x + 2 = 0. * Solution: Here a = 3, b = -7, c = 2. x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(3)(2)}}{2(3)} = \frac{7 \pm \sqrt{49 - 24}}{6} = \frac{7 \pm \sqrt{25}}{6} = \frac{7 \pm 5}{6} x_1 = \frac{12}{6} = 2, \quad x_2 = \frac{2}{6} = \frac{1}{3} The roots are 2 and 1/3.

6. Common Mistakes * Sign of -b: If b is negative, -b is positive. For x^2 - 3x + 2 = 0, the formula starts with +3, not -3. * Fraction line: Ensure the division line extends under the entire term -b \pm \sqrt{\Delta}, not just the square root. * Squaring negative numbers: Remember that (-b)^2 is always positive. For example, (-5)^2 = 25, not -25.

7. Practice Questions 1. Solve x^2 + 6x + 9 = 0 using the quadratic formula. * Answer: x = -3 (repeated root). 2. Solve x^2 - 2x - 4 = 0. Leave in surd form. * Answer: x = 1 \pm \sqrt{5}.

8. Summary / Cheat Sheet * Standard Form: ax^2 + bx + c = 0. * Formula: x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. * Discriminant \Delta = b^2 - 4ac decides root type: positive = two real, zero = one real, negative = two complex.