University:
Glendale Community College
Course:
MAT120 | Intermediate Algebra
Academic year:

2022
Views:

255

Pages:

19
Author:

Deandre Simmons

Section 1.8 Solving Quadratic Equations by Factoring #1-21: Solve each equation. 1) (x-3)(x+2) = 0 x-3 = 0 x = 3 x+2 = 0 x = -2 Solution x = 3, -2 3) (x-1)(x-7) = 0 x-1 = 0 x = 1 x-7 = 0 x = 7 Solution x = 1, 7 5) (3x+12)(2x-8) = 0 3x+12 = 0 3x = -12 x = -4 2x-8 = 0 2x = 8 x = 4 Solution x = 4, -4 7) (2x-9)(3x+10) = 0 2x-9 = 0 2x = 9 x = 9/2 3x+10 = 0 3x = -10 x = -10/3 Solution x = 9/2, -10/3 9) (5x-7)(3x-11) = 0 5x-7 = 0 5x = 7 x = 7/5 3x-11 = 0 3x = 11 x = 11/3 Solution x = 7/5, 11/3 #43-60: Solve each equation. (Remember to factor out the GCF first) 43) 3x³ + 5x² + 2x = 0 x(3x² + 5x + 2) = 0 x(x + 2/3)(x + 1) = 0 x = 0, -2/3, -1 45) 2x² + 2x - 6 = 0 2(x² + x - 3) = 0 2(x + 3/2)(x - 1) = 0 x = -3/2, 1 47) 5x² - 20 = 0 5(x² - 4) = 0 5(x + 2)(x - 2) = 0 5 = 0 (no solution) x + 2 = 0 x = -2 x - 2 = 0 x = 2 Solution x = ±2 49) 3x³ - 15x² + 18x = 0 3x(x² - 5x + 6) = 0 3x(x - 2)(x - 3) = 0 3x = 0 x = 0 x - 2 = 0 x = 2 x - 3 = 0 x = 3 Solution x = 0, 2, 3 51) 3x² - 15x - 18 = 0 3(x² - 5x - 6) = 0 3(x + 1)(x - 6) = 0 3 = 0 (no solution) x + 1 = 0 x = -1 x - 6 = 0 x = 6 Solution x = -1, 6 53) 10x² + 18x + 8 = 0 2(5x² + 9x + 4) = 0 2(x + 4/5)(x + 1) = 0 2 = 0 (no solution) 5x + 4 = 0 5x = -4 x = -4/5 x + 1 = 0 x = -1 Solution x = -4/5, -1