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 - Mid-Chapter Review - Mixed Review: 21

Answer

$25+10\sqrt{6}$

Work Step by Step

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 given expression, $ (\sqrt{15}+\sqrt{10})^2 ,$ is equivalent to \begin{array}{l}\require{cancel} (\sqrt{15})^2+2(\sqrt{15})(\sqrt{10})+(\sqrt{10})^2 \\\\= 15+2\left(\sqrt{15(10)}\right)+10 \\\\= 25+2\left(\sqrt{150}\right) .\end{array} Extracting the factor that is a perfect power of the index results to \begin{array}{l}\require{cancel} 25+2\left(\sqrt{150}\right) \\\\= 25+2\left(\sqrt{25\cdot6}\right) \\\\= 25+2\left(\sqrt{(5)^2\cdot6}\right) \\\\= 25+2(5)\left(\sqrt{6}\right) \\\\= 25+10\sqrt{6} .\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.