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: 35


The answer is as follows: $(7\sqrt13-7\sqrt5)\div8$.

Work Step by Step

We can solve the expression as follows: $= 7\div(\sqrt5+\sqrt13)$ $= 7\div(\sqrt5+\sqrt13)*(\sqrt5-\sqrt13)/(\sqrt5-\sqrt13)$ (Multiply and divide by $(\sqrt5-\sqrt13)$) $= 7*(\sqrt5-\sqrt13)/(\sqrt5^{2}-\sqrt13^{2})$ (since $(a+b)(a-b)=a^{2}-b^{2}$) $= (7\sqrt5-7\sqrt13)\div(5-13)$ $= (7\sqrt5-7\sqrt13)\div(-8)$ $= (7\sqrt13-7\sqrt5)\div8$ (Multiply and divide by -1.)
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