College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter R - Review Exercises: 112

Answer

$\dfrac{23\sqrt{5}}{30}$

Work Step by Step

Using the properties of radicals, the given expression, $ \dfrac{3}{\sqrt{5}}-\dfrac{2}{\sqrt{45}}+\dfrac{6}{\sqrt{80}} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{3}{\sqrt{5}}-\dfrac{2}{\sqrt{9\cdot5}}+\dfrac{6}{\sqrt{16\cdot5}} \\\\= \dfrac{3}{\sqrt{5}}-\dfrac{2}{\sqrt{(3)^2\cdot5}}+\dfrac{6}{\sqrt{(4)^2\cdot5}} \\\\= \dfrac{3}{\sqrt{5}}-\dfrac{2}{3\sqrt{5}}+\dfrac{6}{4\sqrt{5}} .\end{array} Using the $LCD= 12\sqrt{5} ,$ the expression above simplifies to \begin{array}{l}\require{cancel} \dfrac{12(3)-4(2)+3(6)}{12\sqrt{5}} \\\\= \dfrac{36-8+18}{12\sqrt{5}} \\\\= \dfrac{46}{12\sqrt{5}} \\\\= \dfrac{\cancel{2}(23)}{\cancel{2}(6)\sqrt{5}} \\\\= \dfrac{23}{6\sqrt{5}} .\end{array} Rationalizing the denominator results to \begin{array}{l}\require{cancel} \dfrac{23}{6\sqrt{5}}\cdot\dfrac{\sqrt{5}}{\sqrt{5}} \\\\= \dfrac{23\sqrt{5}}{6(5)} \\\\= \dfrac{23\sqrt{5}}{30} .\end{array}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.