Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 8 - Section 8.3 - Operations with Radicals - Exercise Set: 48

Answer

$7\sqrt{6} +6$

Work Step by Step

RECALL: (i) For non-negative real numbers $a$ and $b$, $\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}$ (ii) For all real numbers $a, b,$ and $c$, $a(b+c) = ab+ac$ Use rule (ii) above to obtain: $=\sqrt{6 } \cdot 7 +\sqrt{6} \cdot \sqrt{6} \\=7\sqrt{6}+\sqrt{6} \cdot \sqrt{6}$ Use rule (i) above to obtain: $=7\sqrt{6} +\sqrt{36} =7\sqrt{6} +\sqrt{6^2} =7\sqrt{6} +6$
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.