Intermediate Algebra (6th Edition)

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

Chapter 8 - Review - Page 531: 1

Answer

$x=1,14$

Work Step by Step

$x^{2}-15x+14=0$ $x^{2}-1x-14x+14=0$ $x(x-1)-14(x-1)=0$ $(x-1)(x-14)=0$ $(x-1)=0$ and $(x-14)=0$ $x=1$ and $x=14$
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