College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 1 - Equations and Inequalities - Exercise Set 1.5 - Page 160: 47

Answer

$x=-1,7$

Work Step by Step

To complete the square, add a term that is the square of half the coefficient of $x$ to both sides of the equation. Then use the square root property to simplify both sides, and then isolate $x$. $x^2+6x=7$ $(\frac{6}{2})^2=(3)^2$ $x^2+6x+(3)^2=7+(3)^2$ $x^2+6x+9=7+9$ $x^2+6x+9=16$ $(x+3)^2=16$ $x+3=±4$ $x=-3±4$ $x=-1,7$
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