Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 1 - 1.6 - Other Types of Equations - 1.6 Exercises - Page 128: 46


Solutions: $x=25$ $x=-29$

Work Step by Step

$(x+2)^{\frac{2}{3}}=9~~$ (Cube both sides): $[(x+2)^{\frac{2}{3}}]^3=9^3$ $(x+2)^2=729$ $x+2=±27$ $x=-2±27$ $x=-2+27=25$ or $x=-2-27=-29$ Check the solutions: $x=25$: $(25+2)^{\frac{2}{3}}=(\sqrt[3] {27})^2=3^2=9$ $x=25$: $(-29+2)^{\frac{2}{3}}=(\sqrt[3] {-27})^2=(-3)^2=9$
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