Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 7 - Section 7.5 - Rationalizing Denominators and Numerators of Radical Expressions - Exercise Set - Page 446: 85

Answer

$x=\left\{ -\dfrac{1}{2},6 \right\}$

Work Step by Step

Equating each factor to zero (Zero-Product Property), then the values of $x$ that satisfy the given equation, $ (x-6)(2x+1)=0 ,$ are \begin{array}{l}\require{cancel} x-6=0 \\\\ x=6 ,\\\\\text{OR}\\\\ 2x+1=0 \\\\ 2x=-1 \\\\ x=-\dfrac{1}{2} .\end{array} Hence, $ x=\left\{ -\dfrac{1}{2},6 \right\} .$
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