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: 53

Answer

$3 ( \sqrt 5 - \sqrt 3) $ or $ 3 \sqrt 5 - 3\sqrt 3$

Work Step by Step

$ \frac{6}{\sqrt 5 + \sqrt 3}$ The conjugate of the denominator is $ \sqrt 5 - \sqrt 3 $. Multiply the denominator and numerator by $ \sqrt 5 - \sqrt 3 $, so the simplified denominator will not contain a radical. Therefore, multiply by 1, choosing $\frac{ \sqrt 5 - \sqrt 3 .}{ \sqrt 5 - \sqrt 3 .}$ for 1. $ \frac{6}{\sqrt 5 + \sqrt 3}$ = $ \frac{6}{\sqrt 5 + \sqrt 3}$ $\times$ $\frac{ \sqrt 5 - \sqrt 3 .}{ \sqrt 5 - \sqrt 3 .}$ = $ \frac{6 ( \sqrt 5 - \sqrt 3 ) }{( \sqrt 5 + \sqrt 3 )( \sqrt 5 - \sqrt 3 )}$ $( \sqrt a - \sqrt b)( \sqrt a + \sqrt b)$ = $ (\sqrt a)^{2}$ - $ (\sqrt b)^{2}$. Therefore, $( \sqrt 5 - \sqrt 3)( \sqrt5 +\sqrt 3)$ = $ (\sqrt 5)^{2}$ - $ (\sqrt 3)^{2}$. = $ \frac{6 ( \sqrt 5 - \sqrt 3 ) }{5 -3}$ = $ \frac{6 ( \sqrt 5 - \sqrt 3 ) }{2}$ = $\frac{6}{2} \times ( \sqrt 5 - \sqrt 3 ) $ = $ 3 ( \sqrt 5 - \sqrt 3 ) $ or $ 3\sqrt 5 - 3\sqrt 3$
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