Solving X⁴ - 5x² - 36 = 0 Why Focus On X = ±3

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This article delves into the step-by-step solution of the quartic equation x⁴ - 5x² - 36 = 0, providing a detailed explanation of each step and addressing the critical question: Why do we specifically consider the solutions x = ±3? This exploration aims to provide a clear understanding of the algebraic manipulations involved and the underlying mathematical principles.

Introduction to Quartic Equations

Quartic equations, which are polynomial equations of the fourth degree, can appear daunting at first glance. However, many quartic equations can be solved using techniques similar to those used for quadratic equations. One common approach is to look for opportunities to factor the quartic equation into simpler quadratic factors. This is the strategy we will employ to solve the equation x⁴ - 5x² - 36 = 0. Understanding the nature of quartic equations is crucial as they frequently appear in various fields of mathematics and physics. Solving these equations involves finding the values of x that satisfy the equation, also known as the roots or zeros of the polynomial. The complexity arises from the fourth-degree term, but with strategic factoring, we can reduce it to manageable quadratic equations.

Step-by-Step Solution

Our initial equation is x⁴ - 5x² - 36 = 0. The first key step is to recognize that this equation can be treated as a quadratic equation in disguise. By substituting y = x², we can transform the equation into a more familiar quadratic form. This transformation is a standard technique for dealing with equations that have a specific structure where the powers of the variable are even. The substitution simplifies the algebraic manipulation and makes the equation easier to factor or solve using the quadratic formula. This method is applicable to any equation that can be written in the form ay⁴ + by² + c = 0. By making the appropriate substitution, we can reduce the complexity and apply techniques we already understand.

1. Transforming the Equation

Let's substitute y = x² into the equation. This gives us y² - 5y - 36 = 0. Now we have a standard quadratic equation that we can solve using factoring, completing the square, or the quadratic formula. The goal here is to find the values of y that satisfy this quadratic equation. This transformation is a crucial step in simplifying the original quartic equation. By reducing the degree of the polynomial, we make it easier to find the roots. Recognizing these patterns is a key skill in solving polynomial equations.

2. Factoring the Quadratic Equation

We need to find two numbers that multiply to -36 and add to -5. These numbers are -9 and 4. Therefore, we can factor the quadratic equation as (y - 9)(y + 4) = 0. Factoring is a powerful technique for solving quadratic equations, but it requires us to identify the correct factors. This method relies on understanding the relationship between the coefficients of the quadratic equation and its roots. Practice and familiarity with factoring techniques are essential for efficient problem-solving.

3. Solving for y

Setting each factor equal to zero gives us y - 9 = 0 or y + 4 = 0. Solving these equations gives us y = 9 or y = -4. These are the solutions for the transformed variable y, but we are interested in finding the solutions for x. The values of y represent the squares of the solutions for x, so we need to take the square root to find the values of x.

4. Substituting Back for x

Now we substitute back x² for y. We have x² = 9 or x² = -4. Substituting back allows us to revert to the original variable and find the solutions in terms of x. This step is crucial for completing the solution and answering the original question.

5. Solving for x

For x² = 9, taking the square root of both sides gives us x = ±3. For x² = -4, taking the square root of both sides gives us x = ±2i, where i is the imaginary unit (√-1). The square root operation introduces both positive and negative solutions, which is why we have ±3 and ±2i. This is a fundamental concept in algebra and is essential for finding all possible solutions to the equation.

Why Focus on x = ±3?

The question of why we specifically consider the values x = ±3 arises from the context of real number solutions. While the solutions x = ±2i are valid complex number solutions, the question often implicitly or explicitly seeks real number solutions. In many practical scenarios, particularly in physics and engineering, we are primarily interested in real solutions. Real solutions correspond to points on the number line, while complex solutions involve imaginary numbers and do not have a direct physical interpretation in many contexts. Therefore, when dealing with real-world problems, we often focus on the real solutions of an equation.

Real vs. Complex Solutions

When we solve x² = 9, the solutions x = 3 and x = -3 are real numbers. These values satisfy the original equation and can be represented on the number line. On the other hand, when we solve x² = -4, the solutions x = 2i and x = -2i are complex numbers. Complex numbers involve the imaginary unit i, which is defined as the square root of -1. Complex solutions are just as valid mathematically, but they often represent different types of solutions that may not be relevant in certain applications. For example, in circuit analysis, real solutions might represent physical currents or voltages, while complex solutions might represent impedances or phase shifts.

Practical Applications

In various applications, such as physics, engineering, and computer science, real solutions are often the primary focus. For example, if the equation represents a physical system, such as the position of an object or the current in a circuit, we are typically interested in real-valued solutions that have a direct physical interpretation. Understanding the context of the problem is crucial in determining which solutions are meaningful. While complex solutions have their place in advanced mathematical analysis, real solutions are often the key to solving practical problems.

The Nature of the Question

The original question implicitly directs our attention to the real solutions by the way it is framed. The focus on the factorization and the appearance of x² - 9 = 0 naturally leads us to consider x = ±3 as the primary solutions of interest. The structure of the problem often provides clues about which solutions are most relevant. By recognizing these clues, we can focus our efforts on the most important aspects of the problem.

Detailed Breakdown of the Solution Steps

To further clarify the solution process, let's revisit the key steps:

  1. Initial Equation: We start with the equation x⁴ - 5x² - 36 = 0.
  2. Substitution: We substitute y = x² to obtain the quadratic equation y² - 5y - 36 = 0. This simplifies the equation and allows us to use familiar techniques for solving quadratic equations.
  3. Factoring: We factor the quadratic equation as (y - 9)(y + 4) = 0. This step is crucial for finding the roots of the quadratic equation.
  4. Solving for y: We find the solutions for y as y = 9 and y = -4. These are the roots of the quadratic equation in terms of y.
  5. Substituting Back: We substitute x² back for y to get x² = 9 and x² = -4. This step reverts the equation back to the original variable x.
  6. Solving for x: We solve for x by taking the square root of both sides, which gives us x = ±3 and x = ±2i. These are the solutions to the original quartic equation.

Focus on x² - 9 = 0

The factorization step, specifically the emergence of x² - 9 = 0, is central to understanding why x = ±3 are significant solutions. This factor arises from the factoring process: x⁴ - 5x² - 36 = (x² - 9)(x² + 4) = 0. Setting x² - 9 = 0 leads to x² = 9, and thus x = ±3. The factor x² - 9 is a difference of squares, which is a common pattern in algebra. Recognizing these patterns is essential for efficient problem-solving.

The Role of x² + 4 = 0

While x² + 4 = 0 leads to the complex solutions x = ±2i, it is the x² - 9 = 0 factor that gives us the real solutions. This distinction is important because, in many contexts, we are primarily interested in real solutions. The complex solutions are equally valid from a mathematical standpoint, but they may not have a direct physical interpretation in certain applications.

Conclusion

In summary, the solutions x = ±3 are considered because they are the real number solutions to the quartic equation x⁴ - 5x² - 36 = 0. These solutions arise from the factorization of the equation and represent the values of x that satisfy the equation within the realm of real numbers. While the equation also has complex solutions (x = ±2i), the focus on x = ±3 often stems from a practical or contextual emphasis on real solutions. Understanding the distinction between real and complex solutions is crucial in various fields of mathematics, physics, and engineering. By breaking down the solution step-by-step, we can see how each factor contributes to the final answer and why certain solutions are more relevant in specific contexts. Therefore, when presented with a quartic equation, it is essential to consider both real and complex solutions, but the context of the problem will often dictate which solutions are most meaningful. This comprehensive approach ensures a thorough understanding of the equation and its solutions.