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 1.6 - Page 179: 55

Answer

The solutions are $x=2$ and $x=12$

Work Step by Step

$(x-5)^{2}-4(x-5)-21=0$ Let $u$ be equal to $x-5$ If $u=x-5$, then $u^{2}=(x-5)^{2}$ Rewrite the original equation using the new variable $u$: $u^{2}-4u-21=0$ Solve by factoring: $(u-7)(u+3)=0$ Set both factors equal to $0$ and solve each individual equation for $u$: $u-7=0$ $u=7$ $u+3=0$ $u=-3$ Substitute $u$ back to $x-5$ and solve for $x$: $x-5=7$ $x=7+5$ $x=12$ $x-5=-3$ $x=-3+5$ $x=2$ The solutions are $x=2$ and $x=12$
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