College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 1 - Section 1.2 - Quadratic Equations - 1.2 Assess Your Understanding - Page 102: 62


$\displaystyle x=\frac{-3\pm 2\sqrt{3}}{3}$

Work Step by Step

$3x(x+2)=1$ $3x^{2}+6x=1$ $3x^{2}+6x-1=0$ We solve using the quadratic formula ($a=3,\ b=6,\ c=-1$): $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ $\displaystyle x=\frac{-6\pm\sqrt{6^{2}-4*3*-1}}{2*3}$ $\displaystyle x=\frac{-6\pm\sqrt{36+12}}{6}$ $\displaystyle x=\frac{-6\pm\sqrt{48}}{6}$ $\displaystyle x=\frac{-6\pm\sqrt{16*3}}{6}$ $\displaystyle x=\frac{-6\pm 4\sqrt{3}}{6}$ $\displaystyle x=\frac{-3\pm 2\sqrt{3}}{3}$
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