Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Appendix A - Review - A.11 Complex Numbers - A.11 Assess Your Understanding - Page A94: 3


$6-3\sqrt 3$

Work Step by Step

By multiplying both denominator and nominator by $2-\sqrt{3}$, the given expression simplifies to: $=\dfrac{3}{2+\sqrt{3}} \cdot \dfrac{2-\sqrt{3}}{2-\sqrt{3}}$ $=\dfrac{3(2-\sqrt{3})}{2^2-\sqrt9}$ $=\dfrac{6-3\sqrt{3}}{4-3}$ $=\dfrac{6-3\sqrt{3}}{1}$ $=6-3\sqrt 3$
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.