## College Algebra (6th Edition)

Published by Pearson

# Chapter P - Prerequisites: Fundamental Concepts of Algebra - Exercise Set P.3 - Page 48: 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) $\frac{11}{4} (\sqrt 7 + \sqrt 3)$

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