Algebra 1

Published by Prentice Hall
ISBN 10: 0133500403
ISBN 13: 978-0-13350-040-0

Chapter 10 - Radical Expressions and Equations - 10-4 Solving Radical Equations - Mixed Review - Page 625: 59


$-\frac{2\sqrt 3-4\sqrt 2}{5}$

Work Step by Step

You need to multiply the numerator and denominator by the conjugate which is $\sqrt 3-\sqrt 8$: $\frac{2}{\sqrt 3+\sqrt 8}$$\times$$\frac{\sqrt 3-\sqrt 8}{\sqrt 3-\sqrt 8}$$=$$\frac{2\sqrt 3-2\sqrt 8}{-5}$ Now, simplify the radical: $\frac{2\sqrt 3-2\sqrt 8}{-5}$=$\frac{2\sqrt 3-4\sqrt 2}{-5}$=$-\frac{2\sqrt 3-4\sqrt 2}{5}$
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