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: 10


$x=10$ $x={\dfrac{11}{3}}$

Work Step by Step

$x=\dfrac{-b±{\sqrt {b^{2}-4ac}}}{2a}$ and $3x^{2}-41x+110=0$.Therefore, $a=3$,$b=-41$ and $c=110$: Substitute the values: $x=\dfrac{41±{\sqrt {-41^{2}-4(3)(110)}}}{2(3)}$ =$\dfrac{41±{\sqrt {1681-1320}}}{6}$ =$\dfrac{41± {\sqrt {361}}}{6}$ =$\dfrac{41±19}{6}$ Separate the equation into plus and minus equations: $x=\dfrac{41+19}{6}=10$ $x=\dfrac{41-19}{6}={\dfrac{11}{3}}$
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