Quadratic Equations

Definition: A quadratic equation is an algebraic equation of the form ax² + bx + c = 0, where a ≠ 0 and x is the variable. The highest power of x is 2, so it is called a second-degree equation. The solutions of a quadratic equation are called roots. A quadratic equation has two roots, which can be real, equal, or imaginary.

Types of Quadratic Equations:
1.Pure Quadratic Equation: x² - c = 0 (middle term missing). Example: x² - 16 = 0
2.Complete Quadratic Equation: ax² + bx + c = 0 (all terms present). Example: x² - 5x + 6 = 0
3.Factorable Quadratic Equation: Can be expressed as a product of linear factors. Example: x² - 5x + 6 = (x - 2)(x - 3)
4.Non-Factorable Quadratic Equation: Cannot be factorized easily; solved using formula. Example: x² - 5x - 24 = 0

Properties of Quadratic Equations:
Sum of roots: α + β = -b/a
Product of roots: αβ = c/a
Quadratic equations can have two distinct real roots (Δ > 0), one real double root (Δ = 0), or two complex roots (Δ < 0).

Discriminant (Δ): Δ = b² - 4ac
Δ > 0 → two distinct real roots
Δ = 0 → one real double root
Δ < 0 → two imaginary roots

Methods to Solve Quadratic Equations: Factorization, Completing the Square, Quadratic Formula

11.1 Factorization Method

Concept: Express the quadratic equation as the product of two linear factors, then apply zero-product property.

Steps: Split the middle term if needed. Factor by grouping. Solve each linear factor = 0.

Example 1 Solve x² - 3x + 2 = 0
x² - 3x + 2 = 0
x² - 2x - x + 2 = 0
x(x - 2) - 1(x - 2) = 0
(x - 2)(x - 1) = 0
x - 2 = 0 → x = 2
x - 1 = 0 → x = 1

Example 2 Solve x² + 3x + 2 = 0
x² + 3x + 2 = 0
x² + 2x + x + 2 = 0
x(x + 2) + 1(x + 2) = 0
(x + 2)(x + 1) = 0
x + 2 = 0 → x = -2
x + 1 = 0 → x = -1

Example 3 Solve x² + 4x + 4 = 0
x² + 4x + 4 = 0
x² + 2x + 2x + 4 = 0
x(x + 2) + 2(x + 2) = 0
(x + 2)(x + 2) = 0
x + 2 = 0 → x = -2 (double root)

11.2 Completing the Square Method

Concept: Convert the quadratic equation into a perfect square trinomial.

Steps: Bring constant to the other side. Add (b/2)² to both sides. Factor as a square: (x + b/2)² = … Solve by taking square root.

Example 4 Solve x² - 4x = -4
x² - 4x = -4
x² - 4x + 4 = 0
(x - 2)² = 0
x - 2 = 0 → x = 2

Example 5 Solve x² - 7x + 12 = 0
x² - 7x + 12 = 0
x² - 7x = -12
x² - 7x + (7/2)² = -12 + (7/2)²
(x - 7/2)² = 25/4
x - 7/2 = ±5/2
x = 7/2 + 5/2 = 6
x = 7/2 - 5/2 = 1

Example 6 General: Solve ax² + bx + c = 0
ax² + bx + c = 0
ax² + bx = -c
x² + (b/a)x = -c/a
x² + (b/a)x + (b/2a)² = (b² - 4ac)/4a²
(x + b/2a)² = (b² - 4ac)/4a²
x = [-b ± √(b² - 4ac)] / 2a

11.3 Quadratic Formula Method

Concept: The quadratic formula solves any quadratic equation.
Formula: x = [-b ± √(b² - 4ac)] / 2a

Example 7 Solve x² - 5x - 6 = 0
a = 1, b = -5, c = -6
x = [-(-5) ± √((-5)² - 4·1·(-6))]/2
x = [5 ± √49]/2
x = [5 ± 7]/2
x = (5 + 7)/2 = 6
x = (5 - 7)/2 = -1

Example 8 Solve x² - 5x - 24 = 0
a = 1, b = -5, c = -24
x = [-(-5) ± √((-5)² - 4·1·(-24))]/2
x = [5 ± √121]/2
x = (5 + 11)/2 = 8
x = (5 - 11)/2 = -3

Important Solved Examples

1. Solve (x - 1)(x + 2) = 0
x - 1 = 0 → x = 1
x + 2 = 0 → x = -2

2. Solve x² - x - 12 = 0
x² - x - 12 = 0
x² - 4x + 3x - 12 = 0
x(x - 4) + 3(x - 4) = 0
(x + 3)(x - 4) = 0
x + 3 = 0 → x = -3
x - 4 = 0 → x = 4

3. Solve x² - 7x + 12 = 0
x² - 7x + 12 = 0
x² - 3x - 4x + 12 = 0
x(x - 3) - 4(x - 3) = 0
(x - 4)(x - 3) = 0
x - 4 = 0 → x = 4
x - 3 = 0 → x = 3

4. Solve x² - 4x - 21 = 0
x² - 4x - 21 = 0
x² + 3x - 7x - 21 = 0
x(x + 3) - 7(x + 3) = 0
(x - 7)(x + 3) = 0
x - 7 = 0 → x = 7
x + 3 = 0 → x = -3

5. Solve x² - 10x + 25 = 0
x² - 10x + 25 = 0
x² - 5x - 5x + 25 = 0
x(x - 5) - 5(x - 5) = 0
(x - 5)(x - 5) = 0
Double root: x = 5

6. Solve x² + 2x = 323
x² + 2x - 323 = 0
x² + 19x - 17x - 323 = 0
x(x + 19) - 17(x + 19) = 0
(x - 17)(x + 19) = 0
x - 17 = 0 → x = 17
x + 19 = 0 → x = -19

7. Solve x² - 13x + 42 = 0
x² - 13x + 42 = 0
x² - 6x - 7x + 42 = 0
x(x - 6) - 7(x - 6) = 0
(x - 7)(x - 6) = 0
x = 7, x = 6

8. Solve x² - 30x + 221 = 0
x² - 30x + 221 = 0
x² - 13x - 17x + 221 = 0
x(x - 13) - 17(x - 13) = 0
(x - 17)(x - 13) = 0
x = 17, x = 13

9. Solve x² + 3x + 2 = 0
x² + 3x + 2 = 0
x² + 2x + x + 2 = 0
x(x + 2) + 1(x + 2) = 0
(x + 2)(x + 1) = 0
x = -2, x = -1

10. Solve x² - 5x - 6 = 0 (using formula)
x = [-(-5) ± √((-5)² - 4·1·(-6))]/2
x = [5 ± √49]/2
x = [5 ± 7]/2 → x = 6, x = -1

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