Algebra 1: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281140
ISBN 13: 978-0-13328-114-9

Chapter 9 - Quadratic Functions and Equations - 9-6 The Quadratic Formula and the Discriminant - Practice and Problem-Solving Exercises - Page 586: 9


$x=-3$ $x={\dfrac{5}{4}}$

Work Step by Step

$x=\dfrac{-b±{\sqrt {b^{2}-4ac}}}{2a}$ and $4x^{2}+7x-15=0$.Therefore, $a=4$,$b=7$ and $c=-15$: Substitute the values: $x=\dfrac{-7±{\sqrt {7^{2}-4(4)(-15)}}}{2(4)}$ =$\dfrac{-7±{\sqrt {49-(-240)}}}{8}$ =$\dfrac{-7± {\sqrt {289}}}{8}$ =$\dfrac{-7±17}{8}$ Separate the equation into plus and minus equations: $x=\dfrac{-7-17}{8}=-3$ $x=\dfrac{-7+17}{8}={\dfrac{5}{4}}$
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