Solving X^4 - 5x^2 - 36 = 0 Why X = ±3 Are The Solutions

In mathematics, solving equations is a fundamental skill. This article delves into the step-by-step process of solving the quartic equation $x^4 - 5x^2 - 36 = 0$. We will explore the algebraic manipulations involved in factoring the equation and finding its roots. A particular focus will be given to understanding why the solutions $x = \pm 3$ are valid, while the solutions derived from $x^2 = -4$ are typically disregarded in the context of real number solutions. Understanding this process enhances one's grasp of algebraic problem-solving techniques and the nature of solutions in different number systems.

Step 1: Factoring the Quartic Equation

To solve the equation $x^4 - 5x^2 - 36 = 0$, the initial step involves recognizing the structure of the equation. This equation is a quartic equation, but its form resembles a quadratic equation. We can rewrite it as $(x^2)^2 - 5(x^2) - 36 = 0$. This form suggests that we can treat $x^2$ as a single variable, say $y$, and rewrite the equation as $y^2 - 5y - 36 = 0$. Factoring this quadratic-like equation is the key to finding the solutions.

The objective is to find two numbers that multiply to -36 and add up to -5. These numbers are -9 and +4. Thus, we can factor the quadratic expression in terms of $y$ as $(y - 9)(y + 4) = 0$. Substituting $x^2$ back for $y$, we get $(x^2 - 9)(x^2 + 4) = 0$. This step is crucial because it breaks down the quartic equation into simpler quadratic factors, making it easier to solve for $x$.

Step 2: Solving the Factors

Now that we have factored the equation into $(x^2 - 9)(x^2 + 4) = 0$, we can set each factor equal to zero and solve for $x$. This gives us two equations: $x^2 - 9 = 0$ and $x^2 + 4 = 0$. The first equation, $x^2 - 9 = 0$, is a difference of squares and can be easily solved. The second equation, $x^2 + 4 = 0$, introduces the concept of imaginary numbers, which we will discuss further below.

For $x^2 - 9 = 0$, we can rewrite it as $x^2 = 9$. Taking the square root of both sides gives us $x = \pm 3$. This means that $x$ can be either +3 or -3. These are the real solutions to this part of the equation. The solutions are real because 3 and -3, when squared, equal 9, which satisfies the equation. The plus-minus sign ($\pm$) is important because both the positive and negative roots satisfy the equation.

For $x^2 + 4 = 0$, we can rewrite it as $x^2 = -4$. Taking the square root of both sides gives us $x = \pm \sqrt{-4}$. Since the square root of a negative number is not a real number, we introduce the imaginary unit $i$, where $i = \sqrt{-1}$. Thus, $\sqrt{-4} = \sqrt{4 \times -1} = \sqrt{4} \times \sqrt{-1} = 2i$. Therefore, the solutions to this part of the equation are $x = \pm 2i$, which are complex numbers. These solutions are crucial for a complete understanding of the quartic equation's roots, even though they are not real numbers.

Step 3: Understanding the Solutions x = ±3

The question arises: Why do we specifically focus on the solutions $x = \pm 3$? The answer lies in the context of the problem and the type of solutions we are looking for. In many mathematical contexts, especially in introductory algebra and calculus, we primarily deal with real number solutions. Real numbers are the numbers that can be found on a number line, including both positive and negative numbers, integers, and fractions. When we solve equations, we often seek the real solutions first.

The solutions $x = \pm 3$ are real numbers. When these values are substituted back into the original equation $x^4 - 5x^2 - 36 = 0$, they satisfy the equation. For $x = 3$, we have $3^4 - 5(3^2) - 36 = 81 - 45 - 36 = 0$. For $x = -3$, we have $(-3)^4 - 5((-3)^2) - 36 = 81 - 45 - 36 = 0$. Both values make the equation true, confirming that they are indeed solutions.

However, the solutions $x = \pm 2i$ are imaginary numbers. Imaginary numbers are multiples of the imaginary unit $i$, where $i$ is defined as the square root of -1. In many real-world applications and mathematical contexts, we are interested in real solutions because they represent tangible quantities or points on a graph. Imaginary solutions, while valid solutions to the equation, do not fall within the scope of real number solutions. They exist in a different number system known as complex numbers, which include both real and imaginary parts.

Why Complex Solutions Matter

It is important to note that while we often focus on real solutions, complex solutions are essential in many areas of mathematics and physics. For instance, in electrical engineering, complex numbers are used to describe alternating current circuits. In quantum mechanics, they are fundamental to the wave functions that describe the behavior of particles. Complex solutions provide a complete picture of the roots of a polynomial equation, which is crucial in advanced mathematical analysis and applications.

The Fundamental Theorem of Algebra states that a polynomial equation of degree $n$ has exactly $n$ complex roots (counting multiplicities). In our case, the equation $x^4 - 5x^2 - 36 = 0$ is a quartic equation (degree 4), so it must have four roots. We found two real roots ($x = \pm 3$) and two complex roots ($x = \pm 2i$), which aligns with the theorem. This demonstrates the importance of considering complex solutions for a complete understanding of polynomial equations.

Conclusion

In conclusion, when solving the quartic equation $x^4 - 5x^2 - 36 = 0$, we factor it into $(x^2 - 9)(x^2 + 4) = 0$. This yields the solutions $x = \pm 3$ and $x = \pm 2i$. The solutions $x = \pm 3$ are real numbers and are often the primary focus in many introductory mathematical contexts. The complex solutions $x = \pm 2i$ are equally valid but are typically considered when a complete set of roots or complex number solutions are required. Understanding both types of solutions provides a more comprehensive understanding of the equation and its mathematical properties. The process highlights the importance of algebraic manipulation, factoring, and the recognition of different types of number systems in solving equations.

By focusing on the real solutions $x = \pm 3$, we acknowledge the common requirement in many mathematical problems to find real, tangible solutions. However, the inclusion of the complex solutions $x = \pm 2i$ underscores the richness and completeness of mathematical analysis, where the broader view encompassing complex numbers is often essential for a full understanding of the problem. Therefore, while the focus on $x = \pm 3$ is practical in certain contexts, a comprehensive mathematical understanding necessitates recognizing the existence and significance of complex solutions as well.