Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter R - Review of Basic Concepts - Chapter R Test Prep - Review Exercises - Page 84: 106



Work Step by Step

Bring out the square root of $16$, which is 4, to obtain: $=-4\sqrt{\dfrac{1}{3}}$ Rationalize the denominator by multiplying $3$ to both the numerator and the denominator of the radicand to obtain: $=-4\sqrt{\dfrac{1(3)}{3(3)}} \\=-4\sqrt{\dfrac{3}{3^2}}$ Bring out the square root of the denominator to obtain: $=\color{blue}{-\dfrac{4}{3}\sqrt{3}}$
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