Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 7 - Section 7.5 - Multiplying and Dividing Radical Expressions - 7.5 Exercises - Page 476: 91

Answer

$\sqrt{m}-2$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To rationalize the given radical expression, $ \dfrac{m-4}{\sqrt{m}+2} ,$ multiply the numerator and the denominator by the conjugate of the denominator. Then use special products to simplify the result. $\bf{\text{Solution Details:}}$ Multiplying the numerator and the denominator of the given expression by the conjugate of the denominator, the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{m-4}{\sqrt{m}+2} \cdot\dfrac{\sqrt{m}-2}{\sqrt{m}-2} \\\\= \dfrac{(m-4)(\sqrt{m}-2)}{(\sqrt{m}+2)(\sqrt{m}-2)} .\end{array} Using the product of the sum and difference of like terms which is given by $(a+b)(a-b)=a^2-b^2,$ the expression above is equivalent \begin{array}{l}\require{cancel} \dfrac{(m-4)(\sqrt{m}-2)}{(\sqrt{m})^2-(2)^2} \\\\= \dfrac{(m-4)(\sqrt{m}-2)}{m-4} \\\\= \dfrac{(\cancel{m-4})(\sqrt{m}-2)}{\cancel{m-4}} \\\\= \sqrt{m}-2 .\end{array}
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