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 - 11.3 Studying Solutions of Quadratic Equations - 11.3 Exercise Set - Page 718: 16

Answer

$2$ different imaginary-number solutions that are complex conjugates.

Work Step by Step

$ x^{2}-2x+4=0\qquad$.... $a=1,\ b=-2,\ c=4$ $ b^{2}-4ac\qquad$....substitute $b$ for $-2,\ a$ for $1$ and $c$ for $4$ $=(-2)^{2}-4\cdot 1\cdot 4$ $=4-16$ $=-12$ Since the discriminant is a negative number, there are $2$ different imaginary-number solutions that are complex conjugates.
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