College Algebra (11th Edition)

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

Chapter R - Review Exercises - Page 76: 108



Work Step by Step

Getting the factor that is a perfect root of the index, the given expression, $ \sqrt{\dfrac{2^7y^8}{m^3}} ,$ simplifies to \begin{array}{l}\require{cancel} \sqrt{\dfrac{2^6y^8}{m^2}\cdot\dfrac{2}{m}} \\\\= \sqrt{\left(\dfrac{2^3y^4}{m}\right)^2\cdot\dfrac{2}{m}} \\\\= \dfrac{2^3y^4}{m}\sqrt{\dfrac{2}{m}} .\end{array} Rationalizing the denominator results to \begin{array}{l}\require{cancel} \dfrac{2^3y^4}{m}\sqrt{\dfrac{2}{m}\cdot\dfrac{m}{m}} \\\\= \dfrac{2^3y^4}{m}\sqrt{\dfrac{2m}{(m)^2}} \\\\= \dfrac{2^3y^4}{m}\cdot\dfrac{\sqrt{2m}}{\sqrt{(m)^2}} \\\\= \dfrac{2^3y^4}{m}\cdot\dfrac{\sqrt{2m}}{m} \\\\= \dfrac{2^3y^4\sqrt{2m}}{m^2} \\\\= \dfrac{8y^4\sqrt{2m}}{m^2} .\end{array}
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