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Ch. 1 - Equations and Inequalities
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Not the one you use?Change textbook
Chapter 2, Problem 16

Solve each radical equation in Exercises 11–30. Check all proposed solutions. √(6x + 1) = x - 1

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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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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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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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