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 - Page 50: 123

Answer

In a fraction to rationalize denominator means to replace an irrational number by a rational number in the denominator without changing the value of the fraction. Explanation a). Rationalizing the denominator of $\frac{1}{\sqrt 5}$ = $\frac{\sqrt 5}{ 5}$ b). a). Rationalizing the denominator of $\frac{1}{5 + \sqrt 5}$ = $\frac{{(5 - \sqrt 5)}}{{20}}$

Work Step by Step

In a fraction to rationalize denominator means to replace irrational number by rational number in denominator without changing the value of fraction. Explanation a). Rationalizing the denominator of $\frac{1}{\sqrt 5}$ = . $\frac{1\times{\sqrt 5}}{\sqrt 5\times{\sqrt 5}}$ = $\frac{\sqrt 5}{ 5}$ b). a). Rationalizing the denominator of $\frac{1}{5 + \sqrt 5}$ multiply both nominator and denominator by complex conjugate of $(5 + \sqrt 5)$ = $(5 - \sqrt 5)$ = . $\frac{1\times{(5 - \sqrt 5)}}{(5 + \sqrt 5)\times{(5 - \sqrt 5)}}$ = $\frac{{(5 - \sqrt 5)}}{(5 + \sqrt 5)\times{(5 - \sqrt 5)}}$ by formula ${(a + b){(a - b)}}$ = $a^{2} - b^{2}$ $\frac{{(5 - \sqrt 5)}}{(5 + \sqrt 5)\times{(5 - \sqrt 5)}}$ = $\frac{{(5 - \sqrt 5)}}{5^{2} - \sqrt 5^{2}}$ = $\frac{{(5 - \sqrt 5)}}{{25} - {5}}$ = $\frac{{(5 - \sqrt 5)}}{{20}}$
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