Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

APPENDIX H - Complex Numbers - H Exercises - Page A 64: 23



Work Step by Step

$z^{2}+z+2=0$ We solve using the quadratic formula (a=1, b=1, c=2): $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ $z=\displaystyle \frac{-1\pm\sqrt{1^{2}-4(1)(2)}}{2(1)}$ $=\frac{-1\pm\sqrt{1-8}}{2}$ $=\frac{-1\pm\sqrt{-7}}{2}$ $=\frac{-1\pm\sqrt{-1*7}}{2}$ We use the fact that $\sqrt{-1}=i$: $=-\frac{1}{2}\pm\frac{\sqrt{7}}{2}i$
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