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: 11


$x=-2\pm i\sqrt{3}$

Work Step by Step

Using the properties of equality, the given quadratic equation, $ x^2+4x=-7 ,$ is equivalent to \begin{array}{l}\require{cancel} x^2+4x+7=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{-4\pm\sqrt{4^2-4(1)(7)}}{2(1)} \\\\ x=\dfrac{-4\pm\sqrt{16-28}}{2} \\\\ x=\dfrac{-4\pm\sqrt{-12}}{2} \\\\ x=\dfrac{-4\pm \sqrt{-1}\cdot\sqrt{12}}{2} \\\\ x=\dfrac{-4\pm i\cdot\sqrt{4\cdot3}}{2} \\\\ x=\dfrac{-4\pm i\cdot\sqrt{(2)^2\cdot3}}{2} \\\\ x=\dfrac{-4\pm2i\sqrt{3}}{2} \\\\ x=\dfrac{2(-2\pm i\sqrt{3})}{2} \\\\ x=\dfrac{\cancel{2}(-2\pm i\sqrt{3})}{\cancel{2}} \\\\ x=-2\pm i\sqrt{3} .\end{array}
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