Problem 48 Solve each equation. Express ans... [FREE SOLUTION]

Chapter 7: Problem 48

Solve each equation. Express answers in the form $a+b i$. $$x^{9}-x=0$$

Short Answer

Expert verified

0, 1, -1, i, -i, \frac{\text{鈭�2}}{2} + i\frac{\text{鈭�2}}{2}, \frac{\text{鈭�2}}{2} - i\frac{\text{鈭�2}}{2}, -\frac{\text{鈭�2}}{2} + i\frac{\text{鈭�2}}{2}, -\frac{\text{鈭�2}}{2} - i\frac{\text{鈭�2}}{2}.

Step by step solution

Factor the Equation

First, observe the given equation: \[ x^{9} - x = 0 \] We can factor out an $ x $ from each term: \[ x(x^{8} - 1) = 0 \]

Solve the Factors

Set each factor equal to zero: \[ x = 0 \] and \[ x^{8} - 1 = 0 \]

Solve the Second Factor

Rewrite $ x^{8} - 1 = 0 $ as: \[ x^{8} = 1 \]The solutions are the 8th roots of unity: \[ x = e^{2k\frac{\text{蟺}i}{8}} \] for $ k = 0, 1, 2, ..., 7 $.

Calculate the Solutions

For each $ k \, \text{from} \, 0 \, \text{to} \, 7 $, solve: \[ x = e^{2k\frac{\text{蟺}i}{8}} \]Using Euler's formula $ e^{i\theta} = \text{cos}(\theta) + i \, \text{sin}(\theta) $, find: \[ x = \text{cos}\bigg(\frac{2k\text{蟺}}{8}\bigg) + i \, \text{sin}\bigg(\frac{2k\text{蟺}}{8}\bigg) \]

List All Solutions

Thus, the complete set of solutions in the form $ a + bi $ includes: \[ x = 0, 1, -1, i, -i, \frac{\text{鈭�2}}{2} + i\frac{\text{鈭�2}}{2}, \frac{\text{鈭�2}}{2} - i\frac{\text{鈭�2}}{2}, -\frac{\text{鈭�2}}{2} + i\frac{\text{鈭�2}}{2}, -\frac{\text{鈭�2}}{2} - i\frac{\text{鈭�2}}{2} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

factoring polynomials

Factoring polynomials is a method used to simplify polynomial equations by breaking them down into simpler components, which are called factors. Each factor is a polynomial of lower degree. This can make it easier to solve the equation.

In the original exercise, the equation given is: \[ x^9 - x = 0 \]

We start by factoring out the common factor, which in this case is x:\[ x(x^8 - 1) = 0 \]

Now, we have two factors: x and $ x^8 - 1 $. We solve each factor separately to find:

$ x = 0 $
$ x^8 - 1 = 0 $

The second factor $ x^8 - 1 $ can itself be factored further, but solving it as it stands is manageable by finding the 8th roots of unity.

roots of unity

Roots of unity are solutions to the equation $ x^n = 1 $. They are particularly important in solving polynomial equations with complex numbers. For the given exercise, we are dealing with the 8th roots of unity.

The 8th roots of unity are the 8 complex solutions to the equation: \[ x^8 = 1 \]

These solutions can be expressed in exponential form: \[ x = e^{2k\frac{\text{蟺}i}{8}} \] for $ k = 0, 1, 2, ..., 7 $. The exponential form uses Euler's formula, where: \[ e^{i\theta} = \text{cos}(\theta) + i \text{sin}(\theta) \]

Substituting different values of $ k $, we get:

$ x = e^{0i} = 1 $
$ x = e^{\frac{\text{蟺}i}{4}} = \text{cos}\bigg(\frac{\text{蟺}}{4}\bigg) + i \text{sin}\bigg(\frac{\text{蟺}}{4}\bigg) = \frac{\text{鈭�2}}{2} + i\frac{\text{鈭�2}}{2} $
and so on...

This way, we get 8 unique solutions around the complex unit circle.

complex numbers

Complex numbers are an extension of the real numbers and include imaginary numbers, represented as $ i $ (where $ i^2 = -1 $). They are generally written in the form $ a + bi $.

In the context of the given exercise, we need to express the roots of the polynomial in the form $ a + bi $. For the previously derived 8th roots of unity, we can use Euler's formula to convert them: \[ e^{i\theta} = \text{cos}(\theta) + i \text{sin}(\theta) \]

This results in the following solutions:

$ 1 $
$ \frac{\text{鈭�2}}{2} + i\frac{\text{鈭�2}}{2} $
$ i $
$ -\frac{\text{鈭�2}}{2} + i\frac{\text{鈭�2}}{2} $
$ -1 $
$ -\frac{\text{鈭�2}}{2} - i\frac{\text{鈭�2}}{2} $
$ -i $
$ \frac{\text{鈭�2}}{2} - i\frac{\text{鈭�2}}{2} $

Plus the original root:

$ 0 $

Thus, we have all the roots of the equation $ x^9 - x = 0 $ in the form $ a + bi $.

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91影视

Solve each equation. Express answers in the form \(a+b i\). $$x^{9}-x=0$$

Short Answer

Step by step solution

Factor the Equation

Solve the Factors

Solve the Second Factor

Calculate the Solutions

List All Solutions

Key Concepts

factoring polynomials

roots of unity

complex numbers

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