Textbook Question
Solve each equation in Exercises 15–34 by the square root property.
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 30
Solve each equation in Exercises 15–34 by the square root property. (4x - 1)2 = 16
Verified step by step guidance1
Start by applying the square root property to both sides of the equation. The square root property states that if \((a)^2 = b\), then \(a = \pm \sqrt{b}\). Here, \((4x - 1)^2 = 16\), so take the square root of both sides to get \(4x - 1 = \pm \sqrt{16}\).
Simplify the square root on the right-hand side. Since \(\sqrt{16} = 4\), the equation becomes \(4x - 1 = \pm 4\). This means there are two cases to solve: \(4x - 1 = 4\) and \(4x - 1 = -4\).
Solve the first case, \(4x - 1 = 4\). Add 1 to both sides to isolate the \(4x\) term, resulting in \(4x = 5\). Then divide both sides by 4 to solve for \(x\), giving \(x = \frac{5}{4}\).
Solve the second case, \(4x - 1 = -4\). Add 1 to both sides to isolate the \(4x\) term, resulting in \(4x = -3\). Then divide both sides by 4 to solve for \(x\), giving \(x = \frac{-3}{4}\).
Combine the solutions from both cases. The solutions to the equation are \(x = \frac{5}{4}\) and \(x = \frac{-3}{4}\).
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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 a quadratic equation is in the form (ax + b)^2 = c, then the solutions can be found by taking the square root of both sides. This results in two possible equations: ax + b = √c and ax + b = -√c. This property is essential for solving equations that involve squares.
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Isolating the Variable
Isolating the variable involves rearranging the equation to get the variable on one side and the constants on the other. In the context of the square root property, this often means simplifying the equation to the form (ax + b)^2 = c before applying the square root. This step is crucial for accurately finding the values of the variable.
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Extraneous Solutions
Extraneous solutions are solutions that emerge from the algebraic process but do not satisfy the original equation. When using the square root property, it is important to check each potential solution by substituting it back into the original equation to ensure it is valid. This helps avoid incorrect conclusions drawn from the algebraic manipulation.
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