Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson

Chapter 12 - Exponential Functions and Logarithmic Functions - 12.4 Properties of Logarithmic Functions - 12.4 Exercise Set - Page 810: 77

Answer

$\{-2\pm i\}$

Work Step by Step

$x^{2}+4x+5=0$ $a=1,b=4,c=5$ $b^{2}-4ac=16-20=-4$, negative so the solutions are a complex conjugate pair $x=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}=\frac{-4\pm\sqrt{-4}}{2}$ $=\displaystyle \frac{-4\pm 2i}{2}$ $=\displaystyle \frac{2(-2\pm i)}{2}=-2\pm i$ Solution set = $\{-2\pm i\}$

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