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

Answer

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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