Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 7 - Review - Page 469: 93

Answer

$17\sqrt{2}-15\sqrt{5}$

Work Step by Step

Using the properties of radicals, the expression $ 2\sqrt{50}-3\sqrt{125}+\sqrt{98} $ simplifies to \begin{array}{l} 2\sqrt{25\cdot2}-3\sqrt{25\cdot5}+\sqrt{49\cdot2} \\= 2\cdot5\sqrt{2}-3\cdot5\sqrt{5}+7\sqrt{2} \\= 10\sqrt{2}-15\sqrt{5}+7\sqrt{2} \\= (10\sqrt{2}+7\sqrt{2})-15\sqrt{5} \\= 17\sqrt{2}-15\sqrt{5} .\end{array}
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