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 - Page 68: 82

Answer

$15+10\sqrt{2}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To simplify the given expression, $ (\sqrt{5}+\sqrt{10})^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{5})^2+2(\sqrt{5})(\sqrt{10})+(\sqrt{10})^2 \\\\= 5+2\sqrt{5(10)}+10 \\\\= (5+10)+2\sqrt{50} \\\\= 15+2\sqrt{25\cdot2} \\\\= 15+2\sqrt{(5)^2\cdot2} \\\\= 15+2(5)\sqrt{2} \\\\= 15+10\sqrt{2} .\end{array}
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