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 - Page 439: 64

Answer

$3x-4$

Work Step by Step

Using $(a+b)(a-b)=a^2-b^2$ or the product of the sum and difference of like terms, then, \begin{array}{l} \left( \sqrt[]{3x}+2 \right)\left( \sqrt[]{3x}-2 \right) \\= (\sqrt[]{3x})^2-(2)^2 \\= 3x-4 .\end{array}
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