Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 8 - Radical Functions - 8.3 Multiplying and Dividing Radicals - 8.3 Exercises: 88

Answer

$-40+10\sqrt {7a}-8\sqrt {2}+3\sqrt {14a}$

Work Step by Step

$\sqrt {14a}-(10+\sqrt {8})(4-\sqrt {7a})$ =$\sqrt {14a}-[10(4-\sqrt {7a})+\sqrt {8}(4-\sqrt {7a})]$ =$\sqrt {14a}-(40-10\sqrt {7a}+4\sqrt {8}-\sqrt {8\times7a})$ =$\sqrt {14a}-40+10\sqrt {7a}-4\sqrt {4\times2}+\sqrt {4\times2\times7a}$ =$\sqrt {14a}-40+10\sqrt {7a}-4(2)\sqrt {2}+2\sqrt {14a}$ =$-40+10\sqrt {7a}-8\sqrt {2}+2\sqrt {14a}+\sqrt {14a}$ =$-40+10\sqrt {7a}-8\sqrt {2}+3\sqrt {14a}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.