College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 1 - Section 1.3 - Complex Numbers - 1.3 Exercises: 102

Answer

$(-3+4i)^2+6(-3+4i)+25=0$

Work Step by Step

In order to see that $-3+4i$ is a solution, we plug it in and make sure the equation remains true. $(-3+4i)^2+6(-3+4i)+25=0?$ Expand the left hand side. $9-24i+16i^2-18+24i+25$ Combine like terms using the fact that $i^2=-1$. $(9-16-18+25)+(-24+24)i=0$ $(-3+4i)^2+6(-3+4i)+25=0$
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