College Algebra (10th Edition)

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

Chapter 1 - Review Exercises: 5

Answer

There are no real solutions.

Work Step by Step

We solve: $x(1-x)=6$ $x-x^{2}=6$ $x^{2}-x+6=0$ We solve using the quadratic formula ($a=1,\ b=-1,\ c=6$): $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ However, we note that the discriminant is negative: $b^{2}-4ac=(-1)^{2}-4*1*6=1-24=-23$ Therefore, there are no real solutions.
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