## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 8 - Complex Numbers, Polar Equations, and Parametric Equations - Section 8.4 De Moivre's Theorem: Powers and Roots of Complex Numbers - 8.4 Exercises - Page 383: 49

#### Answer

The three solutions for the value of $x$ are: $x = 1$ $x = \frac{-1 + \sqrt{3}~i}{2}$ $x = \frac{-1 - \sqrt{3}~i}{2}$

#### Work Step by Step

$x^3-1=0$ $(x-1)(x^2+x+1) = 0$ If $(x-1) = 0$, then $x = 1$ We can use the quadratic formula to find the solutions when $(x^2+x+1) = 0$: $x = \frac{-1 \pm \sqrt{1^2-(4)(1)(1)}}{(2)(1)}$ $x = \frac{-1 \pm \sqrt{-3}}{2}$ $x = \frac{-1 \pm \sqrt{3}~i}{2}$ The three solutions for the value of $x$ are: $x = 1$ $x = \frac{-1 + \sqrt{3}~i}{2}$ $x = \frac{-1 - \sqrt{3}~i}{2}$ These three solutions are equivalent to the three solutions found in Exercise 31.

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