College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 1 - Review Exercises: 43

Answer

$\displaystyle \frac{1}{2}\pm\frac{\sqrt{23}}{2}i$

Work Step by Step

$x(1-x)=6$ $x-x^2=6$ $-x^{2}+x-6=0$ We solve using the quadratic formula ($a=-1,\ b=1,\ c=-6$): $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ $x=\frac{-1\pm\sqrt{1^{2}-4*-1*-6}}{2(-1)}$ $x=\frac{-1\pm\sqrt{1-24}}{2(-1)}$ $x=\frac{-1\pm\sqrt{-23}}{-2}$ $x=\frac{-1\pm\sqrt{23}i}{-2}$ $x=\frac{1}{2}\pm\frac{\sqrt{23}}{2}i$
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