Quadratic Equations by Factoring

Example 1: Difference of Perfect Squares

1. Solve: (x^2 - 49 = 0)
   • Difference of Perfect Squares: [x^2 - 49 = (x + 7)(x - 7)]
   • Set each factor to 0:
     • (x + 7 = 0) ⟹ (x = -7)
     • (x - 7 = 0) ⟹ (x = 7)
   • Solutions: (x = -7, 7)

Example 2: Factor by GCF, Then Difference of Squares

1. Solve: (3x^2 - 75 = 0)
   • Greatest Common Factor (GCF): 3
     • [3(x^2 - 25) = 0]
   • Difference of Perfect Squares: [x^2 - 25 = (x + 5)(x - 5)]
   • Set each factor to 0:
     • (x + 5 = 0) ⟹ (x = -5)
     • (x - 5 = 0) ⟹ (x = 5)
   • Solutions: (x = -5, 5)

Example 3: Another Difference of Perfect Squares

1. Solve: (9x^2 - 64 = 0)
   • Difference of Perfect Squares: [9x^2 - 64 = (3x + 8)(3x - 8)]
   • Set each factor to 0:
     • (3x + 8 = 0) ⟹ (x = -\frac{8}{3})
     • (3x - 8 = 0) ⟹ (x = \frac{8}{3})
   • Solutions: (x = -\frac{8}{3}, \frac{8}{3})

Example 4: Factoring a Trinomial (Leading Coefficient = 1)

1. Solve: (x^2 - 2x - 15)
   • Find two numbers that multiply to -15 and add to -2
     • Numbers: -5 and 3
   • Factor: [(x - 5)(x + 3) = 0]
   • Set each factor to 0:
     • (x - 5 = 0) ⟹ (x = 5)
     • (x + 3 = 0) ⟹ (x = -3)
   • Solutions: (x = 5, -3)

Example 5: Factoring a Trinomial (Leading Coefficient ≠ 1)

1. Solve: (8x^2 + 2x - 15)
   • Multiply leading and constant coefficients: (8 \times -15 = -120)
   • Find two numbers that multiply to -120 and add to 2
     • Numbers: 12 and -10
   • Replace 2x with 12x and -10x: [8x^2 + 12x - 10x - 15]
   • Factor by grouping:
     • [4x(2x + 3) - 5(2x + 3)]
   • Factor out common term: [(2x + 3)(4x - 5) = 0]
   • Set each factor to 0:
     • (2x + 3 = 0) ⟹ (x = -\frac{3}{2})
     • (4x - 5 = 0) ⟹ (x = \frac{5}{4})
   • Solutions: (x = -\frac{3}{2}, \frac{5}{4})

Solving by Quadratic Formula

Example 1

1. Solve: (x^2 - 2x - 15 = 0)
   • Quadratic Formula: [x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}]
   • Identify coefficients: (a = 1, b = -2, c = -15)
   • Substitute into formula:
     • [-2 + \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-15)} = \frac{10}{2} = 5]
     • [-2 - \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-15)} = \frac{-6}{2} = -3]
   • Solutions: (x = 5, -3)

Example 2

1. Solve: (8x^2 + 2x - 15 = 0)
   • Quadratic Formula: [x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}]
   • Identify coefficients: (a = 8, b = 2, c = -15)
   • Substitute into formula:
     • [-2 + \sqrt{2^2 - 4 \cdot 8 \cdot (-15)} = \frac{20}{16} = \frac{5}{4}]
     • [-2 - \sqrt{2^2 - 4 \cdot 8 \cdot (-15)} = \frac{-24}{16} = -\frac{3}{2}]
   • Solutions: (x = \frac{5}{4}, -\frac{3}{2})