University:
High School
Course:
High School Mathematics
Academic year:

2022
Views:

70

Pages:

14
Author:

Houston Watkins

CHAPTER 5 Problems and Solutions a) 44 - 2x = x - 10 44 - 2x = x - 10 44 - 2x - x = x - 10 - x 44 - 3x = -10 44 - 3x - 44 = -10 - 44 -3x = -54 -3x/-3 = -54/-3 x = 18 b) √(x + 12) = x (√(x + 12))² = x² x + 12 = x² 0 = x² - x - 12 0 = (x - 4)(x + 3) x = 4, x = -3 x = 4 is the solution. x = -3 is extraneous. c) √(3x - 3) - √(x + 12) = 0 √(3x - 3) = √(x + 12) (√(3x - 3))² = (√(x + 12))² 3x - 3 = x + 12 2x - 3 = 12 2x = 15 x = 15/2 d) (3x + 1)/2 = (-6x - 2)/x Cross multiply x(3x + 1) = 2(-6x - 2) 3x² + x = -12x - 4 3x² + 13x + 4 = 0 (x + 4)(3x + 1) = 0 x = -4, x = -1/3 e) 5^x = 33 Method 1: log(5^x) = log(33) x * log(5) = log(33) x = log(33) / log(5) x = log₅(33) Method 2: Convert exponential equation to logarithmic equation log₅(33) = x f) 11^(6x) = 38 log(11^(6x)) = log(38) 6x * log(11) = log(38) x = log(38) / (6 * log(11)) x = log(38) / log(11^6) 2. What is the extraneous answer for √(2x + 30) = x + 3 Solution: (√(2x + 30))² = (x + 3)² 2x + 30 = x² + 6x + 9 0 = x² + 4x - 21 0 = (x + 7)(x - 3) x = -7, x = 3 Check each solution: √(2(-7) + 30) = -7 + 3 √16 = -4 4 = -4 The square root cannot result in a negative answer. x = -7 is the extraneous solution. 3. f(x) = x² - 10x + 21, g(x) = 4x - 12 a) (f - g)(x) (f - g)(x) = f(x) - g(x) = x² - 10x + 21 - (4x - 12) = x² - 10x + 21 - 4x + 12 = x² - 14x + 33 b) (f + g)(x) (f + g)(x) = f(x) + g(x) = x² - 10x + 21 + 4x - 12 = x² - 6x + 9 c) (f/g)(x) (f/g)(x) = f(x) / g(x) = (x² - 10x + 21) / (4x - 12) = ((x - 7)(x - 3)) / (4(x - 3)) = (x - 7) / 4 20.) A right triangle is drawn with sides labeled 10, 20, and x. 30°-60°-90°: Shorter leg is half the hypotenuse. 45°-45°-90°: Side lengths are congruent Hypotenuse = side * √2 x = 10√2 21.) Describe the transformations. a) 3cos(1/2)(x + π/2) Vertical stretch by a factor of 3, Horizontal stretch by 2, Horizontal translation left π/2 (phase shift = -π/2), Amplitude = 3 Period = 2π/|b| = 2π/(1/2) = 4π Midline: y = 0 b) -2sin(x - π) + 4 Reflection over x-axis, Vertical stretch by a factor of 2, Horizontal translation right π (phase shift = π), Vertical translation up 4. Amplitude = 2 Period = 2π/|b| = 2π/1 = 2π Midline: y = 4 22.) Graph a) sin x A sine wave is graphed with a period of 2π and amplitude of 1. b) cos 2x A cosine wave is graphed with a period of π and amplitude of 1. c) 2sin(x - π/2) A sine wave is graphed with an amplitude of 2 and a phase shift of π/2 to the right.