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 - Mid-Chapter Review - Mixed Review: 22

Answer

$2\sqrt{x-1}$

Work Step by Step

Extracting the factor that is a perfect power of the index results to \begin{array}{l}\require{cancel} \sqrt{25x-25}-\sqrt{9x-9} \\\\= \sqrt{25(x-1)}-\sqrt{9(x-1)} \\\\= \sqrt{(5)^2\cdot (x-1)}-\sqrt{(3)^2\cdot(x-1)} \\\\= 5\sqrt{x-1}-3\sqrt{x-1} \\\\= 2\sqrt{x-1} .\end{array}
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