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: 68

Answer

$128x+26x\sqrt{14}$

Work Step by Step

Using $(a+b)(c+d)=ac+ad+bc+bd$, the given expression, $ \left( 5\sqrt{7x}-\sqrt{2x} \right)\left( 4\sqrt{7x}+6\sqrt{2x} \right) ,$ is equivalent to \begin{array}{l}\require{cancel} 5\sqrt{7x}(4\sqrt{7x})+5\sqrt{7x}(6\sqrt{2x})-\sqrt{2x}( 4\sqrt{7x})-\sqrt{2x}(6\sqrt{2x}) .\end{array} Using the properties of radicals, the expression above simplifies to \begin{array}{l}\require{cancel} 5(4)\sqrt{7x(7x)}+5(6)\sqrt{7x(2x)}-4\sqrt{2x(7x)}-6\sqrt{2x(2x)} \\\\= 20\sqrt{(7x)^2}+30\sqrt{14x^2}-4\sqrt{14x^2}-6\sqrt{(2x)^2} \\\\= 20\sqrt{(7x)^2}+30\sqrt{x^2\cdot 14}-4\sqrt{x^2\cdot 14}-6\sqrt{(2x)^2} \\\\= 20\sqrt{(7x)^2}+30\sqrt{(x)^2\cdot 14}-4\sqrt{(x)^2\cdot 14}-6\sqrt{(2x)^2} \\\\= 20(7x)+30(x)\sqrt{14}-4(x)\sqrt{14}-6(2x) \\\\= 140x+30x\sqrt{14}-4x\sqrt{14}-12x \\\\= 128x+26x\sqrt{14} .\end{array} Note that the variables are assumed to have positive values.
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