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: 69

Answer

$x=\dfrac{1}{2}\pm\dfrac{1}{2}i$

Work Step by Step

$2x^{2}-2x+1=0$ Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. For this equation, $a=2$, $b=-2$ and $c=1$. Substitute the known values in the formula: $x=\dfrac{-(-2)\pm\sqrt{(-2)^{2}-4(2)(1)}}{2(2)}=\dfrac{2\pm\sqrt{4-8}}{4}=...$ $...=\dfrac{2\pm\sqrt{-4}}{4}=\dfrac{2\pm2i}{4}=\dfrac{1}{2}\pm\dfrac{1}{2}i$ The answer is $x=\dfrac{1}{2}\pm\dfrac{1}{2}i$
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