Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 11 - Section 11.3 - Solving Equations by Using Quadratic Methods - Exercise Set - Page 788: 37

Answer

$x=27, 125$

Work Step by Step

$x^{2/3}-8x^{1/3}+15=0$ $y=x^{1/3}$ $x^{2/3}-8x^{1/3}+15=0$ $(x^{1/3})^2-8x^{1/3}+15=0$ $y^2-8y+15=0$ $(y-3)(y-5)=0$ $y-3=0$ $y=3$ $y-5=0$ $y=5$ $y=x^{1/3}$ $3=x^{1/3}$ $3^3=(x^{1/3})^3$ $27=x$ $y=x^{1/3}$ $5=x^{1/3}$ $5^3=(x^{1/3})^3$ $125=x$
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