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: 23

Answer

$x = -1.46~~$ or $~~x = 5.46$

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 $c$: $x^2-4x -8 = 0$ $a = 1$ $b = -4$ $c = -8$ We can use the quadratic formula to find the solutions of the equation: $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ $x = \frac{-(-4) \pm \sqrt{(-4)^2-(4)(1)(-8)}}{(2)(1)}$ $x = \frac{4 \pm \sqrt{16+32}}{2}$ $x = \frac{4 \pm \sqrt{48}}{2}$ $x = \frac{4 - \sqrt{48}}{2}~~$ or $~~x = \frac{4 + \sqrt{48}}{2}$ $x = -1.46~~$ or $~~x = 5.46$
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