Algebra 1

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

Chapter 10 - Radical Expressions and Equations - 10-3 Operations with Radial Expressions - Practice and Problem-Solving Exercises: 32

Answer

$\frac{2\sqrt{6}-2\sqrt{11}}{5}$

Work Step by Step

Multiply using the conjugate. Remember that $(a-b)(a+b)=a^2-b^2$.$\frac{-2}{\sqrt{6}+\sqrt{11}}\cdot \frac{\sqrt{6}-\sqrt{11}}{\sqrt{6}-\sqrt{11}}=\frac{-2\sqrt{6}+2\sqrt{11}}{6-11}=\frac{-2\sqrt{6}+2\sqrt{11}}{-5}=\frac{2\sqrt{6}-2\sqrt{11}}{5}$
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