Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 9 - Section 9.1 - Solving Quadratic Equations by the Square Root Property - Exercise Set: 8

Answer

$x=\pm3\sqrt{3}$

Work Step by Step

Since $x^2=a$ implies $x=\pm\sqrt{a}$ (the Square Root Property), then the solutions of the given equation, $ x^2=27 ,$ are \begin{array}{l}\require{cancel} x=\pm\sqrt{27} \\\\ x=\pm\sqrt{9\cdot3} \\\\ x=\pm\sqrt{(3)^2\cdot3} \\\\ x=\pm3\sqrt{3} .\end{array}
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