Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 10 - Exponents and Radicals - 10.5 Expressions Containing Several Radical Terms - 10.5 Exercise Set - Page 661: 42

Answer

$3+2\sqrt{5}$

Work Step by Step

Using $(a+b)(c+d)=ac+ad+bc+bd$, or the product of 2 binomials, and the properties of radicals, the given expression, $ (4-\sqrt{5})(2+\sqrt{5}) ,$ is equivalent to \begin{array}{l}\require{cancel} 4(2)+4(\sqrt{5})-\sqrt{5}(2)-\sqrt{5}(\sqrt{5}) \\\\= 8+4\sqrt{5}-2\sqrt{5}-\sqrt{5(5)} \\\\= 8+4\sqrt{5}-2\sqrt{5}-\sqrt{(5)^2} \\\\= 8+4\sqrt{5}-2\sqrt{5}-5 \\\\= (8-5)+(4\sqrt{5}-2\sqrt{5}) \\\\= 3+2\sqrt{5} .\end{array}
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