College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 1, Equations and Graphs - Section 1.4 - Solving Quadratic Equations - 1.4 Exercises: 47

Answer

$x\displaystyle \approx0.259$ or $x\displaystyle \approx-0.248$

Work Step by Step

We use the quadratic formula with $a=1$, $b=-0.011$, and $c=-0.064$: $x^{2}-0.011x-0.064=0$ $x=\frac{-(-0.011)\pm\sqrt{(-0.011)^{2}-4(1)(-0.064)}}{2(1)}=\frac{0.011\pm\sqrt{0.000121+0.256}}{2}\approx\frac{0.011\pm 0.506}{2}$ So we have: $x\displaystyle \approx\frac{0.011+0.506}{2}=0.259$ Or: $x\displaystyle \approx\frac{0.011-0.506}{2}=-0.248$
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