College Algebra (10th Edition)

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

Chapter 1 - Section 1.3 - Complex Numbers; Quadratic Equations in the Complex Number System - 1.3 Assess Your Understanding - Page 112: 66


$\displaystyle \frac{3}{13}\pm\frac{2}{13}i$

Work Step by Step

$13x^{2}+1=6x$ $13x^{2}-6x+1=0$ We solve using the quadratic formula ($a=13,\ b=-6,\ c=1$): $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ $x=\frac{-(-6)\pm\sqrt{(-6)^{2}-4(13)(1)}}{2(13)}$ $x=\frac{-(-6)\pm\sqrt{36-52}}{2(13)}$ $x=\frac{+6\pm\sqrt{-16}}{26}$ $x=\frac{6\pm 4i}{26}=\frac{3}{13}\pm\frac{2}{13}i$ Recall, $i=\sqrt{-1}$
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