Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Appendix G - Complex Numbers - G Exercises - Page A56: 22



Work Step by Step

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