## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 8 - Complex Numbers, Polar Equations, and Parametric Equations - Section 8.1 Complex Numbers - 8.1 Exercises - Page 364: 95

#### Answer

The solution set of this problem, written in standard form, is $$\Big\{-\frac{1}{2}\pm\frac{\sqrt3}{2}i\Big\}$$

#### Work Step by Step

$$x^2+1=-x$$ The equation is not in standard form, so first, we need to bring it back to standard form: $$x^2+x+1=0$$ Now the equation is in standard form, the quadratic formula can be applied. $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ As $a=1, b=1, c=1$ $$x=\frac{-1\pm\sqrt{1^2-4\times1\times1}}{2\times1}$$ $$x=\frac{-1\pm\sqrt{1-4}}{2}$$ $$x=\frac{-1\pm\sqrt{-3}}{2}$$ Now we rewrite $\sqrt{-3}=i\sqrt{3}$ $$x=\frac{-1\pm i\sqrt3}{2}$$ The solution set of this problem, written in standard form, is $$\Big\{-\frac{1}{2}\pm\frac{\sqrt3}{2}i\Big\}$$

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