Algebra 1: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281140
ISBN 13: 978-0-13328-114-9

Chapter 10 - Radical Expressions and Equations - 10-2 Simplifying Radicals - Practice and Problem-Solving Exercises - Page 624: 70

Answer

$x=\pm 3\sqrt 2 -3$

Work Step by Step

$x^2+6x-9=0$ $x^2+6x+9=18$ $(x+3)^2=18$ $x+3=\pm \sqrt 18$ $x+3=\pm 3\sqrt 2$ $x=\pm 3\sqrt 2 -3$
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