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: 51

Answer

$7 (\sqrt 5 + 2)$ or $ 7 \sqrt5 + 14 $

Work Step by Step

$ \frac{7}{\sqrt 5 - 2}$ The conjugate of the denominator is $ \sqrt 5 + 2$. Multiply the denominator and numerator by $ \sqrt 5 + 2 $, so the simplified denominator will not contain a radical. Therefore, multiply by 1, choosing $\frac{ \sqrt 5 + 2}{ \sqrt 5 + 2}$ for 1. $ \frac{7}{\sqrt 5 - 2}$ = $ \frac{7}{\sqrt 5 - 2} \times \frac{ \sqrt 5 + 2}{ \sqrt 5 + 2}$ = $\frac{ 7(\sqrt 5 + 2)}{( \sqrt 5 - 2)( \sqrt 5 + 2)}$ $( \sqrt a - \sqrt b)( \sqrt a + \sqrt b)$ = $ (\sqrt a)^{2}$ - $ (\sqrt b)^{2}$. Therefore, $( \sqrt 5 - 2)( \sqrt 5 + 2)$ = $ (\sqrt 5)^{2}$ - $ (2)^{2}$. = $\frac{ 7(\sqrt 5 + 2)}{ (\sqrt 5)^{2} - (2)^{2}}$ = $\frac{ 7(\sqrt 5 + 2)}{ 5 - 4}$ = $\frac{ 7(\sqrt 5 + 2)}{ 1}$ = $7 (\sqrt 5 + 2)$ or $ 7 \sqrt5 + 14 $
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