College Algebra 7th Edition

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

Chapter 1, Equations and Graphs - Chapter 1 Review - Exercises - Page 170: 60



Work Step by Step

$x-4\sqrt{x}=32$ We make the substitution: $u=\sqrt{x}$. The equation then becomes: $u^{2}-4u=32$ $ u^{2}-4u-32=0$ $(u-8)(u+4)=0$ $u-8=0$ or $u+4=0$ $u=8$ or $u=-4$ We convert the $u$ solutions to $x$ solutions: $u=\sqrt{x}$ $x=u^2$ If $u=8$, then: $x=8^2=64$ If $u=-4$, then that would mean $u=-4=\sqrt{x}$ which has no solution. So the only solution to the original equation is $x=64$
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