Textbook Question
Solve each radical equation in Exercises 11–30. Check all proposed solutions. √(6x + 1) = x - 1
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 equation in Exercises 15–34 by the square root property.
Verified step by step guidance1
Start with the given equation: \(5x^2 = 45\).
Divide both sides of the equation by 5 to isolate \(x^2\): \(x^2 = \frac{45}{5}\).
Simplify the right side: \(x^2 = 9\).
Apply the square root property, which states that if \(x^2 = a\), then \(x = \pm \sqrt{a}\): \(x = \pm \sqrt{9}\).
Simplify the square root to find the two possible values for \(x\).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Square Root Property
The square root property states that if x² = k, then x = ±√k. This method is used to solve quadratic equations that can be written in the form x² = a constant. It involves isolating the squared term and then taking the square root of both sides, remembering to include both positive and negative roots.
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Imaginary Roots with the Square Root Property
Isolating the Variable Term
Before applying the square root property, the equation must be manipulated so that the squared term stands alone on one side. This often involves dividing or multiplying both sides of the equation to simplify it. For example, dividing both sides of 5x² = 45 by 5 isolates x².
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05:28Equations with Two Variables
Simplifying Square Roots
After taking the square root of both sides, simplifying the radical expression is necessary. This includes factoring out perfect squares from under the root to write the answer in simplest form. For instance, √9 simplifies to 3, which makes the solution clearer and more exact.
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Related Practice
Textbook Question
In Exercises 1–26, solve and check each linear equation. 4(x + 9) = x
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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Textbook Question
In Exercises 15–26, use graphs to find each set. (- 3, 0) ∩ [- 1, 2]
Textbook Question
Solve each equation. Then state whether the equation is an identity, a conditional equation, or an inconsistent equation. 7(x-4) = x + 2
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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