Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 7 - Section 7.5 - Rationalizing Denominators and Numerators of Radical Expressions - Exercise Set - Page 445: 46

Answer

$-\sqrt{6}-3-2\sqrt{2}-2\sqrt{3}$

Work Step by Step

Rationalizing the denominator of $ \dfrac{\sqrt{3}+\sqrt{4}}{\sqrt{2}-\sqrt{3}} $ results to \begin{array}{l} \dfrac{\sqrt{3}+\sqrt{4}}{\sqrt{2}-\sqrt{3}} \cdot \dfrac{\sqrt{2}+\sqrt{3}}{\sqrt{2}+\sqrt{3}} \\\\= \dfrac{(\sqrt{3})(\sqrt{2})+(\sqrt{3})(\sqrt{3})+(\sqrt{4})(\sqrt{2})+(\sqrt{4})(\sqrt{3})}{(\sqrt{2})^2-(\sqrt{3})^2} \\\\= \dfrac{\sqrt{6}+3+\sqrt{8}+\sqrt{12}}{2-3} \\\\= \dfrac{\sqrt{6}+3+\sqrt{4\cdot2}+\sqrt{4\cdot3}}{-1} \\\\= -\sqrt{6}-3-\sqrt{4\cdot2}-\sqrt{4\cdot3} \\\\= -\sqrt{6}-3-2\sqrt{2}-2\sqrt{3} .\end{array}
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.