Intermediate Algebra (6th Edition)

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

Chapter 9 - Section 9.7 - Common Logarithms, Natural Logarithms, and Change of Base - Exercise Set - Page 586: 72

Answer

$x=-6$ and $x=2$

Work Step by Step

We are given the equation $x^{2}+4x=12$. Subtract 12 from both sides to get all terms on one side. $x^{2}+4x-12=0$ We know that -2 and 6 are factors of 12 and the sum of -2 and 6 is 4, which is the coefficient on the middle term. Therefore, the equation can be factored into the form $(x-2)(x+6)$. Set both terms equal to 0 $x-2=0$ Add 2 to both sides. $x=2$ $x+6=0$ Subtract 6 from both sides. $x=-6$
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