Algebra 1: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281140
ISBN 13: 978-0-13328-114-9

Chapter 10 - Radical Expressions and Equations - Chapter Test - Page 657: 17



Work Step by Step

You are given $\sqrt {k}-8=28$. Solve for k: $\sqrt k-8=28$ $\sqrt {(k)^2} = {36}^{2}$ $k=1296$ Check solution: $\sqrt {1296}-8=28$ $36-8=28$ $28=28$ The solution is $k=1296$
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