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 - 10-3 Operations With Radical Expressions - Practice and Problem-Solving Exercises - Page 629: 32


The answer is $(2\sqrt6-2\sqrt11)\div5$

Work Step by Step

We can simplify the expression as follows: $= -2\div(\sqrt6+\sqrt11)$ $= -2\div(\sqrt6+\sqrt11)*(\sqrt6-\sqrt11)\div(\sqrt6-\sqrt11)$ (It is always preferred to keep the denominator after simplifying the radicals, so multiply and divide by $(\sqrt6-\sqrt11)$) $= -2*(\sqrt6-\sqrt11)\div(\sqrt6)^{2}-(\sqrt11)^{2}$ (Since $(a+b)(a-b)=a^{2}-b^{2}$) $= -2\sqrt6+2\sqrt11\div(6-11)$ $= -2\sqrt6+2\sqrt11\div(-5)$ $= (2\sqrt6-2\sqrt11)\div5$ (Multiply and divide by -1)
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