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: 14


$x=8$ $x={-\dfrac{15}{2}}$

Work Step by Step

$x=\dfrac{-b±{\sqrt {b^{2}-4ac}}}{2a}$ and $2x^{2}-x-120=0$.Therefore, $a=2$,$b=-1$ and $c=-120$ Substitute the values: $x=\dfrac{-(-1)±{\sqrt {-1^{2}-4(2)(-120)}}}{2(2)}$ =$\dfrac{1±{\sqrt {1-(-960)}}}{4}$ =$\dfrac{1± {\sqrt {961}}}{4}$ =$\dfrac{1±31}{4}$ Separate the equation into plus and minus equations: $x=\dfrac{1+31}{4}=8$ $x=\dfrac{1-31}{4}={-\dfrac{15}{2}}$
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