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: 55

Answer

$-\dfrac{\sqrt{14}}{10}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To rationalize the given radical expression, $ -\sqrt{\dfrac{7}{50}} ,$ multiply both the numerator and the denominator by an expression that will make the denominator a perfect power of the index. $\bf{\text{Solution Details:}}$ Converting the radicand as an expression that contains a factor that is a perfect power of the index results to \begin{array}{l}\require{cancel} -\sqrt{\dfrac{7}{25\cdot2}} \\\\= -\sqrt{\dfrac{7}{(5)^2\cdot2}} .\end{array} Multiplying the radicand by an expression equal to $1$ which will make the denominator a perfect power of the index results to \begin{array}{l}\require{cancel} -\sqrt{\dfrac{7}{(5)^2\cdot2}\cdot\dfrac{2}{2}} \\\\= -\sqrt{\dfrac{14}{(5)^2\cdot(2)^2}} .\end{array} Using the Quotient Rule of radicals which is given by $\sqrt[n]{\dfrac{x}{y}}=\dfrac{\sqrt[n]{x}}{\sqrt[n]{y}}{},$ the expression above is equivalent to \begin{array}{l}\require{cancel} -\dfrac{\sqrt{14}}{\sqrt{(5)^2\cdot(2)^2}} \\\\= -\dfrac{\sqrt{14}}{5\cdot2} \\\\= -\dfrac{\sqrt{14}}{10} .\end{array}
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