Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 7 - Summary Exercises - Performing Operations with Radicals and Rational Exponents: 7

Answer

$2+\sqrt{6}-2\sqrt{3}-3\sqrt{2}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To simplify the given radical expression, $ (1-\sqrt{3})(2+\sqrt{6}) ,$ use the FOIL Method and the laws of radicals. $\bf{\text{Solution Details:}}$ Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the expression above is equivalent to\begin{array}{l}\require{cancel} 1(2)+1(\sqrt{6})-\sqrt{3}(2)-\sqrt{3}(\sqrt{6}) \\\\= 2+\sqrt{6}-2\sqrt{3}-\sqrt{3}(\sqrt{6}) .\end{array} Using the Product Rule of radicals which is given by $\sqrt[m]{x}\cdot\sqrt[m]{y}=\sqrt[m]{xy},$ the expression above is equivalent to\begin{array}{l}\require{cancel} 2+\sqrt{6}-2\sqrt{3}-\sqrt{3(6)} \\\\= 2+\sqrt{6}-2\sqrt{3}-\sqrt{18} \\\\= 2+\sqrt{6}-2\sqrt{3}-\sqrt{9\cdot2} \\\\= 2+\sqrt{6}-2\sqrt{3}-\sqrt{(3)^2\cdot2} \\\\= 2+\sqrt{6}-2\sqrt{3}-3\sqrt{2} .\end{array}
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