College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter R - Section R.7 - Radical Expressions - R.7 Exercises: 81

Answer

$11+4\sqrt{6}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To simplify the given expression, $ (\sqrt{3}+\sqrt{8})^2 ,$ use the special product on squaring binomials. $\bf{\text{Solution Details:}}$ Using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the expression above is equivalent to \begin{array}{l}\require{cancel} (\sqrt{3})^2+2(\sqrt{3})(\sqrt{8})+(\sqrt{8})^2 \\\\= 3+2\sqrt{3(8)}+8 \\\\= (3+8)+2\sqrt{24} \\\\= 11+2\sqrt{4\cdot6} \\\\= 11+2\sqrt{(2)^2\cdot6} \\\\= 11+2(2)\sqrt{6} \\\\= 11+4\sqrt{6} .\end{array}
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