College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter P - Prerequisites: Fundamental Concepts of Algebra - Exercise Set P.3 - Page 50: 129


The statement does not make sense since the denominator was rationalized correctly.

Work Step by Step

To confirm adequate rationalization of the exercise, we solve in the following manner: $$\frac{7\sqrt {2}}{\sqrt{6}}$$ = $\frac{7\sqrt {2}}{\sqrt{6}} \times \frac{\sqrt 6}{\sqrt 6}$ = $\frac{7\sqrt {2\times6}}{\sqrt{6^2}}$ =$\frac{7\sqrt {12}}{6}$ =$\frac{7\sqrt {2^2 \times 3}}{6}$ = $\frac{7\times2\sqrt {3}}{6}$ =$\frac{7\sqrt {3}}{3}$
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