Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 10 - Section 10.5 - Rationalizing Numerators and Denominators of Radical Expressions - Exercise Set - Page 719: 77



Work Step by Step

$x^{2}-8x=-12$ $x^{2}-8x+12=0$ $x^{2}-2x-6x+12=0$ $x(x-2)-6(x-2)=0$ $(x-2)(x-6)=0$ $x-2=0$ and $x-6=0$ $x=2$ and $x=6$ The solution set is {$2,6$}.
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.