Algebra 1: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281140
ISBN 13: 978-0-13328-114-9

Chapter 10 - Radical Expressions and Equations - Chapter Test - Page 657: 13


$\frac{2\sqrt 6+3}{5}$

Work Step by Step

We are given: $\frac{\sqrt 6}{4-\sqrt 6}$ Multiplying the equation with $\frac{4+\sqrt 6}{4+\sqrt 6}$ $=\frac{\sqrt 6(4+\sqrt 6)}{16-6}$ $=\frac{4\sqrt 6+6}{10}$ $=\frac{2\sqrt 6+3}{5}$
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