Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 7 - Section 7.4 - Adding, Subtracting, and Multiplying Radical Expressions - Exercise Set: 62

Answer

$10-22\sqrt{3}$

Work Step by Step

Using $(a+b)(c+d)=ac+ad+bc+bd$, the given expression, $ \left( \sqrt{6}-4\sqrt{2} \right)\left( 3\sqrt{6}+\sqrt{2} \right) ,$ is equivalent to \begin{array}{l}\require{cancel} \sqrt{6}(3\sqrt{6})+\sqrt{6}(\sqrt{2})-4\sqrt{2}( 3\sqrt{6})-4\sqrt{2}(\sqrt{2}) .\end{array} Using the properties of radicals, the expression above simplifies to \begin{array}{l}\require{cancel} 3\sqrt{6(6)}+\sqrt{6(2)}-4(3)\sqrt{2(6)}-4\sqrt{2(2)} \\\\= 3\sqrt{36}+\sqrt{12}-12\sqrt{12}-4\sqrt{4} \\\\= 3\sqrt{36}+\sqrt{4\cdot3}-12\sqrt{4\cdot3}-4\sqrt{4} \\\\= 3\sqrt{(6)^2}+\sqrt{(2)^2\cdot3}-12\sqrt{(2)^2\cdot3}-4\sqrt{(2)^2} \\\\= 3(6)+2\sqrt{3}-12(2)\sqrt{3}-4(2) \\\\= 18+2\sqrt{3}-24\sqrt{3}-8 \\\\= (18-8)+(2\sqrt{3}-24\sqrt{3}) \\\\= 10-22\sqrt{3} .\end{array}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.