Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 8 - Sections 8.1-8.3 - Integrated Review - Summary on Solving Quadratic Equations: 10

Answer

$x=-3\pm\sqrt{5}$

Work Step by Step

Multiplying both sides by $2$, the given quadratic equation, $ \dfrac{1}{2}x^2+3x+2=0 ,$ is equivalent to \begin{array}{l}\require{cancel} x^2+6x+4=0 .\end{array} Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, the solutions of the quadratic equation above are \begin{array}{l}\require{cancel} x=\dfrac{-6\pm\sqrt{6^2-4(1)(4)}}{2(1)} \\\\ x=\dfrac{-6\pm\sqrt{36-16}}{2} \\\\ x=\dfrac{-6\pm\sqrt{20}}{2} \\\\ x=\dfrac{-6\pm\sqrt{4\cdot5}}{2} \\\\ x=\dfrac{-6\pm\sqrt{(2)^2\cdot5}}{2} \\\\ x=\dfrac{-6\pm2\sqrt{5}}{2} \\\\ x=\dfrac{2(-3\pm\sqrt{5})}{2} \\\\ x=\dfrac{\cancel{2}(-3\pm\sqrt{5})}{\cancel{2}} \\\\ x=-3\pm\sqrt{5} .\end{array}
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