Solving $a + A^2 + A^3 + A^4 - 2 = 0$ Find $\frac{a^4 + 2}{a}$ Value

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Introduction

In this article, we will delve into solving the given equation $a + a^2 + a^3 + a^4 - 2 = 0$ and subsequently determine the value of the expression $\frac{a^4 + 2}{a}$. This problem involves algebraic manipulation and a keen understanding of polynomial equations. We will explore various techniques, including factoring and substitution, to arrive at the solution. This discussion falls under the category of mathematics, specifically algebra, and it aims to provide a comprehensive step-by-step solution that is both clear and insightful.

Understanding the Problem

To begin, let's restate the problem clearly. We are given a fourth-degree polynomial equation: $a + a^2 + a^3 + a^4 - 2 = 0$. Our primary goal is to find the value of the expression $\frac{a^4 + 2}{a}$. This requires us to first find the possible values of $a$ that satisfy the given equation, and then substitute these values into the expression to obtain the final answer. The challenge lies in the fact that solving quartic equations can be complex, and we need to employ strategic methods to simplify the process. Understanding the nature of the polynomial and its potential roots is crucial for solving this problem effectively. We will break down the equation and the expression into manageable parts, making use of algebraic techniques to find the solution.

Factoring and Simplifying the Equation

The given equation is $a + a^2 + a^3 + a^4 - 2 = 0$. To solve this, we can attempt to factor the polynomial. Rearranging the terms, we get $a^4 + a^3 + a^2 + a - 2 = 0$. By observation, we can see that $a = 1$ is a solution because $1^4 + 1^3 + 1^2 + 1 - 2 = 1 + 1 + 1 + 1 - 2 = 2$. Therefore, $(a - 1)$ is a factor of the polynomial. Now, we can perform polynomial long division or synthetic division to divide $a^4 + a^3 + a^2 + a - 2$ by $(a - 1)$. The result of the division is $a^3 + 2a^2 + 3a + 2$. So, the equation can be rewritten as $(a - 1)(a^3 + 2a^2 + 3a + 2) = 0$. We now need to solve the cubic equation $a^3 + 2a^2 + 3a + 2 = 0$. Factoring this cubic equation is not straightforward, but we can look for rational roots using the Rational Root Theorem. The possible rational roots are $\pm 1$ and $\pm 2$. We already know $a = 1$ is not a root of the cubic equation. Let's test $a = -1$: $(-1)^3 + 2(-1)^2 + 3(-1) + 2 = -1 + 2 - 3 + 2 = 0$. Thus, $a = -1$ is a root, and $(a + 1)$ is a factor. Dividing $a^3 + 2a^2 + 3a + 2$ by $(a + 1)$ gives us $a^2 + a + 2$. Therefore, the cubic equation can be factored as $(a + 1)(a^2 + a + 2) = 0$. Now, we have factored the original equation as $(a - 1)(a + 1)(a^2 + a + 2) = 0$. To find the remaining roots, we need to solve the quadratic equation $a^2 + a + 2 = 0$.

Solving the Quadratic Equation

We have the quadratic equation $a^2 + a + 2 = 0$. To find the roots, we can use the quadratic formula, which is given by:

$a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In our equation, $a = 1$, $b = 1$, and $c = 2$. Plugging these values into the quadratic formula, we get:

$a = \frac{-1 \pm \sqrt{1^2 - 4(1)(2)}}{2(1)}$

$a = \frac{-1 \pm \sqrt{1 - 8}}{2}$

$a = \frac{-1 \pm \sqrt{-7}}{2}$

Since the discriminant ($b^2 - 4ac$) is negative, the roots are complex numbers. Thus, the roots of the quadratic equation are:

$a = \frac{-1 + i\sqrt{7}}{2}$ and $a = \frac{-1 - i\sqrt{7}}{2}$

So, the four roots of the original equation $a^4 + a^3 + a^2 + a - 2 = 0$ are $a = 1$, $a = -1$, $a = \frac{-1 + i\sqrt{7}}{2}$, and $a = \frac{-1 - i\sqrt{7}}{2}$. Now that we have the roots, we can move on to finding the value of the expression $\frac{a^4 + 2}{a}$.

Evaluating the Expression $\frac{a^4 + 2}{a}$

We need to find the value of $\frac{a^4 + 2}{a}$ for each root of the equation. Let's start with the real roots, $a = 1$ and $a = -1$.

Case 1: $a = 1$

When $a = 1$, the expression becomes:

$\frac{1^4 + 2}{1} = \frac{1 + 2}{1} = \frac{3}{1} = 3$

Case 2: $a = -1$

When $a = -1$, the expression becomes:

$\frac{(-1)^4 + 2}{-1} = \frac{1 + 2}{-1} = \frac{3}{-1} = -3$

Now, let's consider the complex roots $a = \frac{-1 + i\sqrt{7}}{2}$ and $a = \frac{-1 - i\sqrt{7}}{2}$. Evaluating the expression for these roots is more complex, but we can use the original equation to simplify the calculation. From the given equation, $a + a^2 + a^3 + a^4 - 2 = 0$, we can express $a^4$ as:

$a^4 = 2 - a - a^2 - a^3$

Now, substitute this expression for $a^4$ into the expression $\frac{a^4 + 2}{a}$:

$\frac{a^4 + 2}{a} = \frac{(2 - a - a^2 - a^3) + 2}{a} = \frac{4 - a - a^2 - a^3}{a}$

$\frac{a^4 + 2}{a} = \frac{4}{a} - 1 - a - a^2$

Now we can evaluate this expression for the complex roots.

Case 3: $a = \frac{-1 + i\sqrt{7}}{2}$

Let $a = \frac{-1 + i\sqrt{7}}{2}$. We need to calculate $\frac{4}{a} - 1 - a - a^2$. First, let's find $\frac{1}{a}$:

$\frac{1}{a} = \frac{2}{-1 + i\sqrt{7}} = \frac{2(-1 - i\sqrt{7})}{(-1 + i\sqrt{7})(-1 - i\sqrt{7})} = \frac{-2 - 2i\sqrt{7}}{1 + 7} = \frac{-2 - 2i\sqrt{7}}{8} = \frac{-1 - i\sqrt{7}}{4}$

So, $\frac{4}{a} = -1 - i\sqrt{7}$.

Now, we need to find $a^2$:

$a^2 = \left(\frac{-1 + i\sqrt{7}}{2}\right)^2 = \frac{(-1)^2 + 2(-1)(i\sqrt{7}) + (i\sqrt{7})^2}{4} = \frac{1 - 2i\sqrt{7} - 7}{4} = \frac{-6 - 2i\sqrt{7}}{4} = \frac{-3 - i\sqrt{7}}{2}$

Now, we can substitute these values into the expression $\frac{4}{a} - 1 - a - a^2$:

$\frac{4}{a} - 1 - a - a^2 = (-1 - i\sqrt{7}) - 1 - \left(\frac{-1 + i\sqrt{7}}{2}\right) - \left(\frac{-3 - i\sqrt{7}}{2}\right)$

$= -1 - i\sqrt{7} - 1 + \frac{1}{2} - \frac{i\sqrt{7}}{2} + \frac{3}{2} + \frac{i\sqrt{7}}{2}$

$= -2 + \frac{1}{2} + \frac{3}{2} - i\sqrt{7} = -2 + 2 - i\sqrt{7} = -i\sqrt{7}$

Case 4: $a = \frac{-1 - i\sqrt{7}}{2}$

Let $a = \frac{-1 - i\sqrt{7}}{2}$. We follow a similar process as in Case 3. First, let's find $\frac{1}{a}$:

$\frac{1}{a} = \frac{2}{-1 - i\sqrt{7}} = \frac{2(-1 + i\sqrt{7})}{(-1 - i\sqrt{7})(-1 + i\sqrt{7})} = \frac{-2 + 2i\sqrt{7}}{1 + 7} = \frac{-2 + 2i\sqrt{7}}{8} = \frac{-1 + i\sqrt{7}}{4}$

So, $\frac{4}{a} = -1 + i\sqrt{7}$.

Now, we need to find $a^2$:

$a^2 = \left(\frac{-1 - i\sqrt{7}}{2}\right)^2 = \frac{(-1)^2 + 2(-1)(-i\sqrt{7}) + (-i\sqrt{7})^2}{4} = \frac{1 + 2i\sqrt{7} - 7}{4} = \frac{-6 + 2i\sqrt{7}}{4} = \frac{-3 + i\sqrt{7}}{2}$

Now, we can substitute these values into the expression $\frac{4}{a} - 1 - a - a^2$:

$\frac{4}{a} - 1 - a - a^2 = (-1 + i\sqrt{7}) - 1 - \left(\frac{-1 - i\sqrt{7}}{2}\right) - \left(\frac{-3 + i\sqrt{7}}{2}\right)$

$= -1 + i\sqrt{7} - 1 + \frac{1}{2} + \frac{i\sqrt{7}}{2} + \frac{3}{2} - \frac{i\sqrt{7}}{2}$

$= -2 + \frac{1}{2} + \frac{3}{2} + i\sqrt{7} = -2 + 2 + i\sqrt{7} = i\sqrt{7}$

Final Values

We have found the values of the expression $\frac{a^4 + 2}{a}$ for all the roots of the given equation:

• For $a = 1$, $\frac{a^4 + 2}{a} = 3$
• For $a = -1$, $\frac{a^4 + 2}{a} = -3$
• For $a = \frac{-1 + i\sqrt{7}}{2}$, $\frac{a^4 + 2}{a} = -i\sqrt{7}$
• For $a = \frac{-1 - i\sqrt{7}}{2}$, $\frac{a^4 + 2}{a} = i\sqrt{7}$

Conclusion

In this detailed analysis, we successfully solved the equation $a + a^2 + a^3 + a^4 - 2 = 0$ and found its four roots. These roots include two real values, $a = 1$ and $a = -1$, and two complex values, $a = \frac{-1 + i\sqrt{7}}{2}$ and $a = \frac{-1 - i\sqrt{7}}{2}$. We then evaluated the expression $\frac{a^4 + 2}{a}$ for each of these roots. The corresponding values are $3$, $-3$, $-i\sqrt{7}$, and $i\sqrt{7}$. This problem showcases the application of various algebraic techniques, including factoring, the Rational Root Theorem, the quadratic formula, and complex number arithmetic. Understanding these methods is crucial for solving polynomial equations and evaluating algebraic expressions efficiently. By breaking down the problem into smaller, manageable parts, we were able to systematically arrive at the solution. This comprehensive approach highlights the importance of careful analysis and methodical execution in mathematical problem-solving. The combination of real and complex solutions underscores the richness and complexity of polynomial algebra. This exercise not only provides a solution to the specific problem but also reinforces fundamental algebraic principles, making it a valuable learning experience. The ability to handle both real and complex roots is essential in advanced mathematics and engineering applications, where such equations frequently arise. Through this detailed walkthrough, readers can gain a deeper appreciation for the intricacies of algebraic manipulations and the beauty of mathematical solutions.