Introduction to Solving a Quartic Equation

The quartic equation (x^4 - x^3 - x^2 - x - 1 0) is a polynomial equation of degree 4. This article will guide us through the process of solving this equation and understanding the properties of its roots. We will explore the concept of roots on the unit circle in the complex plane and identify the roots through both algebraic and geometric approaches.

Algebraic Transformation and Solving for (u)

Let's start by manipulating the given quartic equation. Dividing both sides by (x^2), we get:

[x^2 - frac{1}{x^2} - x - frac{1}{x} - 1 0]

Next, let's set (u x frac{1}{x}). Note that:

[u^2 x^2 2 frac{1}{x^2}]

By substituting (u^2 - 2) for (x^2 frac{1}{x^2}), we can rewrite the equation as:

[u^2 - 2 - u - 1 0]

Simplifying further:

[u^2 - u - 3 0]

Using the quadratic formula to solve for (u):

[u frac{-(-1) pm sqrt{1 12}}{2} frac{1 pm sqrt{5}}{2}]

Thus, the values of (u) are:

(u_1 frac{1 sqrt{5}}{2}) (u_2 frac{1 - sqrt{5}}{2})

Finding the Roots of the Original Equation

Given the original equation (x^4 - x^3 - x^2 - x - 1 0), we can express the values of (x) in terms of (u x frac{1}{x}). Using the quadratic formula for (x frac{1}{x} u):

[x frac{1}{x} u]

Rearranging the quadratic equation:

[x^2 - ux 1 0]

Applying the quadratic formula:

[x frac{u pm sqrt{u^2 - 4}}{2}]

Substituting the values of (u_1) and (u_2) into the quadratic formula:

(x_1 frac{1 sqrt{5}}{4} pm frac{sqrt{5sqrt{5}-5}}{4}) (x_2 frac{1 sqrt{5}}{4} pm frac{sqrt{5}}{4}) (x_3 frac{1 - sqrt{5}}{4} pm frac{sqrt{5sqrt{5}-5}}{4}) (x_4 frac{1 - sqrt{5}}{4} pm frac{sqrt{5}}{4})

These are the four complex roots of the quartic equation.

Exploring the Unit Circle in the Complex Plane

The roots of the original quartic equation are the fifth roots of 1, except for 1 itself. In the complex plane, these roots form a regular pentagon, centered at the origin. Each root is a fifth root of unity, which means each root is of the form (e^{2pi i k / 5}) for (k 0, 1, 2, 3, 4).

Since (a) is a root of the quartic equation (excluding 1), the reciprocal (frac{1}{a}) is simply the reflection of (a) across the real axis. Thus, when we add these two roots:

[a frac{1}{a} 2cos(72^circ) -2cos(36^circ)]

The figure consists of similar triangles, confirming the solution through geometric reasoning.

Conclusion

In summary, the roots of the quartic equation (x^4 - x^3 - x^2 - x - 1 0) can be found using algebraic manipulation and the properties of the unit circle. Understanding the geometric properties of the roots in the complex plane provides a deeper insight into the solutions, especially in relation to the fifth roots of unity.