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the value of the rac condition. Working Example 3: Solve an Equation With T Solve the radical equation √2x + 5 = 2√2x + 1, x ≥ 0. Chea lution

the value of the rac condition. Working Example 3: Solve an Equation With T Solve the radical equation √2x + 5 = 2√2x + 1, x ≥ 0. Chea lution

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Therefore, the you c) The radicand 2-m must be positive for the 2-m=0 2-m+mz0+m 2zm Written another way, m 2. So, the value of m must the value of the radical equation to be a real number. The so- condition. Working Example 3: Solve an Equation With Twe Solve the radical equation √2x + 5 = 2√2x + 1, x ≥ 0. Check- Solution √2x + 5 = 2√√2x + 1 Square both sides. 36x²- (√2x + 5)² = (2√2x + 1)² _)(- = (- 2x + 5 = 4(2x) + 2x + 5 = 8x + 4√2x + 1 2x + 5-8x-1= 8x-8x + 4√2x + 1-1 -6x + 4 = 4√2x (-6x + 4)² = (4√2x)² -)( -) = (- + 16 = 32x -)( +1 214 MHR Chapter 5 978-0-07-073882-9
Transcribed Image Text:Therefore, the you c) The radicand 2-m must be positive for the 2-m=0 2-m+mz0+m 2zm Written another way, m 2. So, the value of m must the value of the radical equation to be a real number. The so- condition. Working Example 3: Solve an Equation With Twe Solve the radical equation √2x + 5 = 2√2x + 1, x ≥ 0. Check- Solution √2x + 5 = 2√√2x + 1 Square both sides. 36x²- (√2x + 5)² = (2√2x + 1)² _)(- = (- 2x + 5 = 4(2x) + 2x + 5 = 8x + 4√2x + 1 2x + 5-8x-1= 8x-8x + 4√2x + 1-1 -6x + 4 = 4√2x (-6x + 4)² = (4√2x)² -)( -) = (- + 16 = 32x -)( +1 214 MHR Chapter 5 978-0-07-073882-9
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