College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 1 - Quiz - Sections 1.1-1.4 - Page 114: 9

Answer

$x=\left\{ -\sqrt{29},\sqrt{29} \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ x^2-29=0 ,$ convert it to the form $x^2=c.$ Then use the Square Root Principle. $\bf{\text{Solution Details:}}$ Using the properties of equality, the equation above is equivalent to \begin{array}{l}\require{cancel} x^2=29 .\end{array} Taking the square root of both sides (Square Root Principle), the equation above is equivalent to \begin{array}{l}\require{cancel} x=\pm\sqrt{29} .\end{array} The solutions are $ x=\left\{ -\sqrt{29},\sqrt{29} \right\} .$
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