College Algebra (11th Edition)

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

Chapter R - Review Exercises - Page 76: 106



Work Step by Step

Getting the factor that is a perfect root of the index, the given expression, $ -\sqrt{\dfrac{16}{3}} ,$ simplifies to \begin{array}{l}\require{cancel} -\sqrt{16\cdot\dfrac{1}{3}} \\\\= -\sqrt{(4)^2\cdot\dfrac{1}{3}} \\\\= -4\sqrt{\dfrac{1}{3}} .\end{array} Rationalizing the denominator results to \begin{array}{l}\require{cancel} -4\sqrt{\dfrac{1}{3}\cdot\dfrac{3}{3}} \\\\= -4\sqrt{\dfrac{3}{(3)^2}} \\\\= -4\cdot\dfrac{\sqrt{3}}{\sqrt{(3)^2}} \\\\= -4\cdot\dfrac{\sqrt{3}}{3} \\\\= -\dfrac{4\sqrt{3}}{3} .\end{array}
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