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 - Page 123: 48


$x=1.25$ or $x=1.2$

Work Step by Step

We use the quadratic formula with $a=1$, $b=-2.45$, and $c=1.5$: $x^{2}-2.450x+1.500=0$ $x=\displaystyle \frac{-(-2.45)\pm\sqrt{(-2.45)^{2}-4(1)(1.5)}}{2(1)}=\frac{2.45\pm\sqrt{6.0025-6}}{2}=\frac{2.45\pm\sqrt{0.0025}}{2}=\frac{2.450\pm 0.05}{2}$. So we have: $x=\displaystyle \frac{2.45+0.05}{2}=1.25$ Or: $x=\displaystyle \frac{2.45-0.05}{2}=1.2$
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