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: 20

Answer

$x=32$

Work Step by Step

Add 5 on both sides of the equation to obtain: $\sqrt[3]{4x-3}=5$ RECALL: For any real number $a$, $(\sqrt[3]{a})^3=a$ Cube both sides then use the rule above to simplify: $(\sqrt[3]{4x-3})^3 = 5^3 \\4x-3=125$ Add 3 on both sides to obtain: $4x = 125+3 \\4x=128$ Divide 4 on both sides to obtain: $x=\frac{128}{4} \\x=32$
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