College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter P - Prerequisites: Fundamental Concepts of Algebra - Exercise Set P.3: 54

Answer

$\frac{11 (\sqrt 7 +\sqrt 3 )}{4}$ (or) $\frac{11}{4} (\sqrt 7 + \sqrt 3)$

Work Step by Step

$ \frac{11}{(\sqrt 7 - \sqrt 3)}$ The conjugate of the denominator is $ \sqrt 7 + \sqrt 3 $. Multiply the denominator and numerator by $ \sqrt 7 + \sqrt 3 $, so the simplified denominator will not contain a radical. Therefore, multiply by 1, choosing $\frac{ \sqrt 7 + \sqrt 3 .}{ \sqrt 7 + \sqrt 3 .}$ for 1. $ \frac{11}{(\sqrt 7 - \sqrt 3)}$ = $ \frac{11}{(\sqrt 7 - \sqrt 3)}$ $\times$ $\frac{ \sqrt 7 + \sqrt 3 .}{ \sqrt 7 + \sqrt 3 .}$ = $ \frac{11 ( \sqrt 7 + \sqrt 3 ) }{( \sqrt 7 - \sqrt 3 )( \sqrt 7 + \sqrt 3 )}$ $( \sqrt a - \sqrt b)( \sqrt a + \sqrt b)$ = $ (\sqrt a)^{2}$ - $ (\sqrt b)^{2}$. Therefore, $( \sqrt 7 - \sqrt 3)( \sqrt7 +\sqrt 3)$ = $ (\sqrt7)^{2}$ - $ (\sqrt 3)^{2}$. = $ \frac{11 ( \sqrt7 + \sqrt 3 ) }{(\sqrt7)^{2} - (\sqrt 3)^{2}}$ = $ \frac{11 ( \sqrt7 + \sqrt 3 ) }{ 7- 3}$ = $ \frac{11 ( \sqrt7 + \sqrt 3 ) }{ 4}$ (or) $\frac{11}{4} (\sqrt 7 + \sqrt 3)$
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.