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

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 11 - Quadratic Functions and Equations - 11.2 The Quadratic Formula - 11.2 Exercise Set - Page 713: 23


$x=-2+\sqrt{2}i$ or $x=-2-\sqrt{2}i$

Work Step by Step

$ x^{2}+4x+6=0\qquad$... solve with the Quadractic formula. $a=1,\ b=4,\ c=6$ $ x=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\qquad$... substitute $b$ for $4, \ a$ for $1$ and $c$ for $6$. $ x=\displaystyle \frac{-4\pm\sqrt{(4)^{2}-4\cdot(6)\cdot 1}}{2\cdot 1}\qquad$... simplify. $x=\displaystyle \frac{-4\pm\sqrt{16-24}}{2}$ $x=\displaystyle \frac{-4\pm\sqrt{-8}}{2}$ $ x=\displaystyle \frac{-4+\sqrt{-1\cdot 4\cdot 2}}{}\qquad$... write in terms of $i$. ($\sqrt{-1}=i$) $x=\displaystyle \frac{-4\pm 2\sqrt{2}i}{2}$ $ x=-2\pm\sqrt{2}i\qquad$... the symbol $\pm$ indicates two solutions. $x=-2+\sqrt{2}i$ or $x=-2-\sqrt{2}i$
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