Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 4 Quadratic Functions and Factoring - 4.7 Complete the Square - 4.7 Exercises - Skill Practice - Page 289: 61

Answer

$x=\frac{-b\pm \sqrt b^2-4c}{2}$

Work Step by Step

Given: $x^2+bx+c=0$ $(x^2+bx)+c=0$ $(x^2+2\times\frac{b}{2}\times x+(\frac{b}{2})^2)-(\frac{b}{2})^2+c=0$ $(x+\frac{b}{2})^2+\frac{4c-b^2}{4}=0$ $(x+\frac{b}{2})^2=\frac{b^2-4c}{4}$ $x+\frac{b}{2}=\pm\sqrt \frac{b^2-4c}{4}$ $x+\frac{b}{2}=\pm \frac{\sqrt b^2-4c}{2}$ $x=-\frac{b}{2}\pm \frac{\sqrt b^2-4c}{2}$ $x=\frac{-b\pm \sqrt {b^2-4c}}{2}$
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