Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.6 - Complex Numbers - 1.6 Exercises - Page 64: 66


$x=3\pm i$

Work Step by Step

$x^{2}-6x+10=0$ Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. For this equation, $a=1$, $b=-6$ and $c=10$. Substitute the known values in the formula: $x=\dfrac{-(-6)\pm\sqrt{(-6)^{2}-4(1)(10)}}{2(1)}=\dfrac{6\pm\sqrt{36-40}}{2}=...$ $...=\dfrac{6\pm\sqrt{-4}}{2}=\dfrac{6\pm2i}{2}=3\pm i$ The answer is $x=3\pm i$.
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