Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 8 - Section 8.2 - The Quadratic Formula - 8.2 Exercises: 8

Answer

$x=\left\{ \dfrac{-3-\sqrt{17}}{4},\dfrac{-3+\sqrt{17}}{4} \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To find the solutions of the given equation, $ 2x^2+3x-1=0 ,$ use the Quadratic Formula. $\bf{\text{Solution Details:}}$ Using the form $ax^2+bx+c=0,$ the quadratic equation above has $a= 2 , b= 3 , c= -1 .$ Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then \begin{array}{l}\require{cancel} x=\dfrac{-3\pm\sqrt{3^2-4(2)(-1)}}{2(2)} \\\\ x=\dfrac{-3\pm\sqrt{9+8}}{4} \\\\ x=\dfrac{-3\pm\sqrt{17}}{4} .\end{array} The solutions are \begin{array}{l}\require{cancel} x=\dfrac{-3-\sqrt{17}}{4} \\\\\text{OR}\\\\ x=\dfrac{-3+\sqrt{17}}{4} .\end{array} Hence, $ x=\left\{ \dfrac{-3-\sqrt{17}}{4},\dfrac{-3+\sqrt{17}}{4} \right\} .$
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