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: 16


$x=\dfrac{1\pm \sqrt{13}}{4}$

Work Step by Step

Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, the solutions of the given quadratic equation, $ 4x^2-2x-3=0 ,$ are \begin{array}{l}\require{cancel} x=\dfrac{-(-2)\pm\sqrt{(-2)^2-4(4)(-3)}}{2(4)} \\\\ x=\dfrac{2\pm\sqrt{4+48}}{8} \\\\ x=\dfrac{2\pm\sqrt{52}}{8} \\\\ x=\dfrac{2\pm\sqrt{4\cdot13}}{8} \\\\ x=\dfrac{2\pm\sqrt{(2)^2\cdot13}}{8} \\\\ x=\dfrac{2\pm2\sqrt{13}}{8} \\\\ x=\dfrac{2(1\pm \sqrt{13})}{8} \\\\ x=\dfrac{\cancel{2}(1\pm \sqrt{13})}{\cancel{2}(4)} \\\\ x=\dfrac{1\pm \sqrt{13}}{4} .\end{array}
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