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: 67

Answer

$x=-\dfrac{1}{2}\pm\dfrac{\sqrt{3}}{2}i$

Work Step by Step

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