Intermediate Algebra (6th Edition)

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

Chapter 9 - Section 9.3 - Exponential Functions - Exercise Set - Page 560: 56


$x=2$ and $x=9$

Work Step by Step

We are given that $18=11x-x^{2}$. We can first subtract 18 from both sides of the equation to get all terms on one side. $-x^{2}+11x-18=0$ Next, we can multiply each term on both sides by -1 so the term $x^{2}$ will have a positive coefficient. $x^{2}-11x+18=0$ We know that 2 and 9 are factors of 18 and the sum of -2 and -9 is equal to $-2+(-9)=-11$ (which is the coefficient attached to the middle term). Therefore, we can factor the existing polynomial into $(x-2)(x-9)=0$ Set both terms in parentheses equal to 0. $x-2=0$ Add 2 to both sides. $x=2$ $x-9=0$ Add 9 to both sides. $x=9$
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