Elementary Geometry for College Students (7th Edition) Clone

Published by Cengage
ISBN 10: 978-1-337-61408-5
ISBN 13: 978-1-33761-408-5

Appendix A - A.5 - The Quadratic Formula and Square Root Properties - Exercises - Page 562: 26

Answer

$x = 0.13~~$ or $~~x = 3.87$

Work Step by Step

A quadratic equation can be written in this form: $ax^2 + bx+c = 0$ where $a,b,$ and $c$ are real numbers and $a \neq 0$ We can determine the values of $a, b,$ and $c$: $2x^2 = 8x-1$ $2x^2 -8x+1 = 0$ $a = 2$ $b = -8$ $c = 1$ We can use the quadratic formula to find the solutions of the equation: $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ $x = \frac{-(-8) \pm \sqrt{(-8)^2-(4)(2)(1)}}{(2)(2)}$ $x = \frac{8 \pm \sqrt{64-8}}{4}$ $x = \frac{8 \pm \sqrt{56}}{4}$ $x = \frac{8 - \sqrt{56}}{4}~~$ or $~~x = \frac{8 + \sqrt{56}}{4}$ $x = 0.13~~$ or $~~x = 3.87$
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