Textbook Question
Including a 10.5% hotel tax, your room in San Diego cost \$216.58 per night. Find the nightly cost before the tax was added.
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Ch. 1 - Equations and Inequalities
Blitzer8th EditionCollege AlgebraISBN: 9780136970514
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Table of contents
- Ch. P - Fundamental Concepts of Algebra311
- Ch. 1 - Equations and Inequalities398
- Ch. 2 - Functions and Graphs407
- Ch. 3 - Polynomial and Rational Functions306
- Ch. 4 - Exponential and Logarithmic Functions247
- Ch. 5 - Systems of Equations and Inequalities173
- Ch. 6 - Matrices and Determinants143
- Ch. 7 - Conic Sections124
- Ch. 8 - Sequences, Induction, and Probability251
Chapter 2, Problem 16
Solve each radical equation in Exercises 11–30. Check all proposed solutions. √(6x + 1) = x - 1
Verified step by step guidance1
Start by isolating the radical expression on one side of the equation, which is already done: \(\sqrt{6x + 1} = x - 1\).
To eliminate the square root, square both sides of the equation: \(\left(\sqrt{6x + 1}\right)^2 = (x - 1)^2\).
Simplify both sides: \(6x + 1 = (x - 1)^2\).
Expand the right side using the formula \((a - b)^2 = a^2 - 2ab + b^2\): \(6x + 1 = x^2 - 2x + 1\).
Rearrange the equation to set it equal to zero by moving all terms to one side: \(0 = x^2 - 2x + 1 - 6x - 1\), then simplify the right side.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Radical Equations
Radical equations involve variables inside a root, such as a square root. To solve them, isolate the radical expression and then eliminate the root by raising both sides to the power that corresponds to the root, typically squaring for square roots.
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Expanding Radicals
Extraneous Solutions
When solving radical equations, squaring both sides can introduce solutions that do not satisfy the original equation. These are called extraneous solutions and must be checked by substituting back into the original equation to verify their validity.
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06:00Categorizing Linear Equations
Domain Restrictions
The expression inside a square root must be non-negative to produce real number solutions. Additionally, the right side of the equation must be consistent with the domain, such as ensuring the expression equals or exceeds zero when isolated, to avoid invalid solutions.
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3:51Domain Restrictions of Composed Functions
Related Practice
Textbook Question
Solve each equation in Exercises 15–34 by the square root property.
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Textbook Question
In Exercises 1–26, solve and check each linear equation. 4(x + 9) = x
900
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Textbook Question
Graph each equation in Exercises 13 - 28. Let x = - 3, - 2, - 1, 0, 1, 2, 3
y = x + 2
876
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Textbook Question
Solve each equation in Exercises 15–34 by the square root property.
858
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Textbook Question
Graph each equation in Exercises 13 - 28. Let x = - 3, - 2, - 1, 0, 1, 2, 3
y = x - 2
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