College Algebra (10th Edition)

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

Chapter 1 - Section 1.2 - Quadratic Equations - 1.2 Assess Your Understanding - Page 102: 67


$x=3.47$ or $x=0.63$

Work Step by Step

We are given: $x^{2}-4.1x+2.2=0$ We solve using the quadratic formula ($a=1,\ b=-4.1,\ c=2.2$): $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ $x=\frac{-(-4.1)\pm\sqrt{(-4.1)^{2}-4*1*2.2}}{2*1}$ $x=\frac{4.1\pm\sqrt{16.81-8.8}}{2}$ $x=\frac{4.1\pm\sqrt{8.01}}{2}$ $x=3.47$ or $x=0.63$
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