College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter P, Prerequisites - Chapter P Review - Exercises: 122

Answer

$x=\pm 8$

Work Step by Step

We solve the equation: $x^{2/3}-4=0$ $(x^{1/3})^{2}=0+4=4$ $x^{1/3}=\pm\sqrt{4}=\pm 2$ $x=(\pm2)^3=\pm 8$
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