University:
University of California, Berkeley
Course:
MATH 56 Linear Algebra
Academic year:

2020
Views:

248

Pages:

8
Author:

larbiwxcvbn

**Quadratics** x² - 3x = 0 → x(x-3) = 0 → x = 0, 3 x² - 3x = -2 → x² - 3x + 2 = 0 → (x-1)(x-2) = 0 → x = 1, 2 More generally ax² + bx + c, a, b, c fixed real numbers, a ≠ 0. How to solve? 1. Factoring 2. Quadratic Formula 3. Completing the square Last method is least used but useful, is what creates 2 ax^2 + bx + c = 0 x^2 + (b/a)x + c/a = 0 x^2 + (b/a)x = -c/a x^2 + (b/a)x + (b/2a)^2 = (b/2a)^2 - c/a (x + b/2a)^2 = (b^2 - 4ac)/4a^2 x + b/2a = ±√(b^2 - 4ac)/√(4a^2) x = -b/2a ± √(b^2 - 4ac)/2a x = (-b ± √(b^2 - 4ac))/2a Quadratic Formula b²-4ac: discriminant If b²-4ac < 0, no real solutions. Cleanly factorable if a, b, c integers? If b²-4ac ≥ 0 with b²-4ac being a perfect square. Ex: x³ + 3x² + 2x = 0 x(x² + 3x + 2) = 0 x(x + 2)(x + 1) = 0 x = -2, -1, 0 Ex: x³ + 3x² + 2x + 6 = 0 x²(x + 3) + 2(x + 3) = 0 (x² + 2)(x + 3) = 0 ↓ never is 0 so x = -3 only Ex: x³ - 5x² + 7x - 3 = 0 Factored last time: (x-1)(x²-4x+3) = 0 (x-1)²(x-3) = 0 x = 1, 3 In this, don't do: x³ - 5x² + 7x = 3 x(x² - 5x + 7) = 3 Why? a * b = 0 → a = 0 or b = 0 a * b = 3 → a = 3 or b = 3? No!!! That's why that doesn't help (x+1)(x+2) = 0 → x = -2, -1 (x+1)(x+2) = 5 → x² + 3x + 2 = 5 → x² + 3x - 3 = 0 So these look similar but in first factored helps, in second it doesn't.