College Algebra (6th Edition)

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

Chapter 2 - Functions and Graphs - Exercise Set 2.7 - Page 311: 102

Answer

$y = 3 \frac{+}{}\sqrt{13}$

Work Step by Step

$$y^2 - 6y - 4 =0$$ Since $(\frac{b}{2a})^2$ is $(\frac{-6}{2})^2 = -3^2 = 9$ for this case, we can re-write the original equation as follows: $$y^2 - 6y + 9 - 9 - 4 = 0$$ $$(y-3)^2 -13 = 0$$ $$y-3 =\frac{+}{}\sqrt{13}$$ $$y = 3 \frac{+}{}\sqrt{13}$$
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