Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter P - Section P.7 - Equations - Concept and Vocabulary Check - Page 105: 16


Fill the blank with: $9$

Work Step by Step

$x^{2}+6x$ is a part of $\quad$ $ x^{2}+2\cdot x\cdot\fbox{$3$} +\fbox{$3^{2}$}=(x+3)^{2}$ so we add $9$ (that is, $3^{2}$) to both sides to complete the square on the LHS. We obtain $x^{2}+6x+9=7+9$ $(x+3)^{2}=16$, from which, by the square root property, follows: $x+3=\pm 4$ $x=-3\pm 4$ Solution set = $\{-7,+1\}$ Fill the blank with: $9$
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