Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole
ISBN 10:
ISBN 13:

Appendix A - A.5 - The Quadratic Formula and Square Root Properties - Exercises - Page 555: 25

Answer

$x = -0.92~~$ or $~~x = 1.52$

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 $c$: $5x^2 = 3x+7$ $5x^2 -3x-7 = 0$ $a = 5$ $b = -3$ $c = -7$ We can use the quadratic formula to find the solutions of the equation: $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ $x = \frac{-(-3) \pm \sqrt{(-3)^2-(4)(5)(-7)}}{(2)(5)}$ $x = \frac{3 \pm \sqrt{9+140}}{10}$ $x = \frac{3 \pm \sqrt{149}}{10}$ $x = \frac{3 - \sqrt{149}}{10}~~$ or $~~x = \frac{3 + \sqrt{149}}{10}$ $x = -0.92~~$ or $~~x = 1.52$
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