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 - Page 518: 7


$x=\left\{ \dfrac{-2-\sqrt{2}}{2},\dfrac{-2+\sqrt{2}}{2} \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To find the solutions of the given equation, $ 2x^2+4x+1=0 ,$ use the Quadratic Formula. $\bf{\text{Solution Details:}}$ Using the form $ax^2+bx+c=0,$ the quadratic equation above has $a= 2 , b= 4 , c= 1 .$ Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then \begin{array}{l}\require{cancel} x=\dfrac{-4\pm\sqrt{4^2-4(2)(1)}}{2(2)} \\\\ x=\dfrac{-4\pm\sqrt{16-8}}{4} \\\\ x=\dfrac{-4\pm\sqrt{8}}{4} \\\\ x=\dfrac{-4\pm\sqrt{4\cdot2}}{4} \\\\ x=\dfrac{-4\pm\sqrt{(2)^2\cdot2}}{4} \\\\ x=\dfrac{-4\pm2\sqrt{2}}{4} \\\\ x=\dfrac{2(-2\pm\sqrt{2})}{4} \\\\ x=\dfrac{\cancel2(-2\pm\sqrt{2})}{\cancel2(2)} \\\\ x=\dfrac{-2\pm\sqrt{2}}{2} .\end{array} The solutions are \begin{array}{l}\require{cancel} x=\dfrac{-2-\sqrt{2}}{2} \\\\\text{OR}\\\\ x=\dfrac{-2+\sqrt{2}}{2} .\end{array} Hence, $ x=\left\{ \dfrac{-2-\sqrt{2}}{2},\dfrac{-2+\sqrt{2}}{2} \right\} .$
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