To solve each of these quadratic equations, we can either apply the quadratic formula or factor them when possible. Let's go through each equation one by one.
Equation: \( x^2 - 10x - 11 = 0 \)
- Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 1, b = -10, c = -11 \)
- \( b^2 - 4ac = 100 + 44 = 144 \)
- \( x = \frac{10 \pm 12}{2} \)
- Solutions: \( x = 11 \) and \( x = -1 \)
- Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Equation: \( x^2 - 9x - 10 = 0 \)
- Factorable form: \( (x - 10)(x + 1) = 0 \)
- Solutions: \( x = 10 \) and \( x = -1 \)
Equation: \( x^2 + 8x - 9 = 0 \)
- Using the quadratic formula:
- Here, \( a = 1, b = 8, c = -9 \)
- \( b^2 - 4ac = 64 + 36 = 100 \)
- \( x = \frac{-8 \pm 10}{2} \)
- Solutions: \( x = 1 \) and \( x = -9 \)
- Using the quadratic formula:
Equation: \( x^2 + 9x + 20 = 0 \)
- Factorable form: \( (x + 5)(x + 4) = 0 \)
- Solutions: \( x = -5 \) and \( x = -4 \)
Equation: \( x^2 - 15x + 36 = 0 \)
- Factorable form: \( (x - 12)(x - 3) = 0 \)
- Solutions: \( x = 12 \) and \( x = 3 \)
Equation: \( x^2 + 5x - 14 = 0 \)
- Using the quadratic formula:
- Here, \( a = 1, b = 5, c = -14 \)
- \( b^2 - 4ac = 25 + 56 = 81 \)
- \( x = \frac{-5 \pm 9}{2} \)
- Solutions: \( x = 2 \) and \( x = -7 \)
- Using the quadratic formula:
Equation: \( x^2 - 7x - 30 = 0 \)
- Factorable form: \( (x - 10)(x + 3) = 0 \)
- Solutions: \( x = 10 \) and \( x = -3 \)
Summary of Solutions:
- \( x^2 - 10x - 11 = 0 \) → \( x = 11, -1 \)
- \( x^2 - 9x - 10 = 0 \) → \( x = 10, -1 \)
- \( x^2 + 8x - 9 = 0 \) → \( x = 1, -9 \)
- \( x^2 + 9x + 20 = 0 \) → \( x = -5, -4 \)
- \( x^2 - 15x + 36 = 0 \) → \( x = 12, 3 \)
- \( x^2 + 5x - 14 = 0 \) → \( x = 2, -7 \)
- \( x^2 - 7x - 30 = 0 \) → \( x = 10, -3 \)