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 - Page 592: 68



Work Step by Step

RECALL: (i) $(a+b)(a-b) = a^2-b^2$ (ii) $\sqrt{a}^2=a, a \ge0$ Use rule (i) above to obtain: $=\sqrt{11}^2-6^2 \\=\sqrt{11}^2-36$ Use rule (ii) above to obtain: $=11-36 \\=-25$
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.