Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 7 - Section 7.6 - Radical Equations - Exercise Set - Page 559: 17

Answer

$x=8$

Work Step by Step

RECALL: For any real number $a$, $(\sqrt[3]{a})^3=a$ Cube both sides then use the rule above to simplify: $(\sqrt[3]{2x+11})^3 = 3^3 \\2x+11=27$ Subtract 11 on both sides to obtain: $2x = 27-11 \\2x=16$ Divide 2 on both sides to obtain: $x=8$
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