Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 10 - Exponents and Radicals - Study Summary - Practice Exercises - Page 691: 10

Answer

$-5\sqrt{2}$

Work Step by Step

Extracting the factor that is a perfect power of the index, then \begin{array}{l}\require{cancel} 5\sqrt{8}-3\sqrt{50} \\\\= 5\sqrt{4\cdot2}-3\sqrt{25\cdot2} \\\\= 5\sqrt{(2)^2\cdot2}-3\sqrt{(5)^2\cdot2} \\\\= 5(2)\sqrt{2}-3(5)\sqrt{2} \\\\= 10\sqrt{2}-15\sqrt{2} .\end{array} Combining the like radicals results to \begin{array}{l}\require{cancel} 10\sqrt{2}-15\sqrt{2} \\\\= (10-15)\sqrt{2} \\\\= -5\sqrt{2} .\end{array}
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