Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 1 - Equations and Inequalities - 1.4 Quadratic Equations - 1.4 Exercises: 27

Answer

$\color{blue}{\left\{-3\sqrt3, 3\sqrt3\right\}}$

Work Step by Step

Add $x^2$ to both sides to obtain: $27=x^2 \\x^2=27$ RECALL: If $x^2=a$, then taking the square root of both sides gives $x = \pm \sqrt{a}$. Take the square root of both sides of the given equation to obtain: $\sqrt{x^2}=\pm \sqrt{27} \\x =\pm \sqrt{9(3)} \\x=\pm\sqrt{3^2(3)} \\x=\pm 3\sqrt{3}$ Thus, the solution set is $\color{blue}{\left\{-3\sqrt3, 3\sqrt3\right\}}$.
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