Solving 9x^4 - 2x^2 - 7 = 0 With U-Substitution A Step-by-Step Guide

In the realm of algebra, solving quartic equations, those with a highest degree of four, can often seem daunting. However, with the strategic application of u-substitution, we can transform these complex equations into more manageable quadratic forms. This article delves into the step-by-step solution of the quartic equation 9x^4 - 2x^2 - 7 = 0, showcasing the power and elegance of u-substitution in simplifying polynomial equations.

Understanding Quartic Equations and the Power of u-Substitution

A quartic equation is a polynomial equation of the fourth degree, generally expressed as ax^4 + bx^2 + c = 0. Directly solving such equations can be intricate, but when the equation lacks cubic and linear terms (terms with x^3 and x), a clever substitution can significantly simplify the process. This is where u-substitution shines.

The essence of u-substitution lies in replacing a complex expression within the equation with a single variable, 'u' in this case. This transformation aims to convert the quartic equation into a quadratic equation, which we can then solve using familiar methods such as factoring, the quadratic formula, or completing the square. Once we find the values of 'u', we reverse the substitution to determine the solutions for the original variable, 'x'.

The equation at hand, 9x^4 - 2x^2 - 7 = 0, perfectly fits the profile for u-substitution due to the absence of x^3 and x terms. This allows us to transform it into a solvable quadratic equation, making the solution process significantly more streamlined. This method is very important in understanding and solving such problems in mathematics. The application of u-substitution is not just a trick; it is a powerful technique rooted in the fundamental principles of algebraic manipulation and equation solving. By simplifying the equation's structure, we unveil its underlying nature and make it amenable to standard solution methods. Furthermore, mastering u-substitution enhances our problem-solving toolkit, equipping us to tackle a broader range of polynomial equations with greater confidence and efficiency. In the subsequent sections, we will dissect the solution process, elucidating each step and highlighting the elegance of this method. This will not only provide the solutions to the specific equation but also impart a valuable problem-solving skill applicable to numerous mathematical contexts.

Step-by-Step Solution Using u-Substitution

1. Introduce the Substitution: Let u = x^2. This substitution is the cornerstone of our approach, transforming the quartic equation into a quadratic one. By replacing x^2 with 'u', we effectively reduce the degree of the equation, making it solvable using standard techniques. The choice of u = x^2 is strategic, as it directly addresses the structure of the quartic equation, which contains only even powers of x. This clever substitution allows us to rewrite the equation in terms of 'u', paving the way for simplification and solution. The act of substitution is a fundamental tool in mathematical problem-solving, allowing us to reframe problems in more manageable forms. In this case, it transforms the problem from the realm of quartics to the more familiar territory of quadratics. This is where we leverage our understanding of simpler equations to tackle more complex ones. Understanding the motivation behind this substitution is crucial; it's not just a random step but a deliberate act of simplification that leverages the specific form of the equation.

2. Rewrite the Equation in Terms of u: Substituting u into the original equation, we get 9u^2 - 2u - 7 = 0. This is a classic quadratic equation, a far cry from the original quartic form. The transformation highlights the power of u-substitution; a complex equation has been elegantly reduced to a familiar form. The rewritten equation is now amenable to standard quadratic solution techniques, such as factoring, the quadratic formula, or completing the square. This step underscores the strategic advantage of u-substitution - it transforms a difficult problem into a manageable one. The transition from quartic to quadratic is not merely a cosmetic change; it represents a significant reduction in complexity. By working with the quadratic equation in 'u', we can leverage our existing knowledge and skills to find the values of 'u', which will then lead us to the solutions for 'x'. This step is a testament to the power of mathematical abstraction and transformation. The ability to reframe a problem in a different context is a key skill in mathematical problem-solving.

3. Solve the Quadratic Equation for u: The quadratic equation 9u^2 - 2u - 7 = 0 can be solved by factoring. We seek two numbers that multiply to (9)(-7) = -63 and add up to -2. These numbers are -9 and 7. We rewrite the middle term and factor by grouping:9u^2 - 9u + 7u - 7 = 0 9u(u - 1) + 7(u - 1) = 0 (9u + 7)(u - 1) = 0 Setting each factor to zero gives us:9u + 7 = 0 => u = -7/9 u - 1 = 0 => u = 1 Thus, the solutions for u are u = -7/9 and u = 1. Factoring is a powerful technique for solving quadratic equations, but it requires a keen eye for number relationships. The ability to identify the correct factors often comes with practice and a solid understanding of number theory. This particular quadratic equation lends itself well to factoring, but other methods, such as the quadratic formula, could also be employed. The crucial aspect is to find the values of 'u' that satisfy the equation, as these values will then be used to find the solutions for the original variable 'x'. The process of factoring involves breaking down the quadratic expression into a product of two linear expressions. This is a fundamental skill in algebra and is applicable in many different contexts. The solutions for 'u' represent the points where the quadratic function crosses the u-axis, and these points are crucial in understanding the behavior of the function.

4. Substitute Back to Find x: Now we substitute back x^2 for u and solve for x:

   • For u = -7/9:x^2 = -7/9 x = ±√(-7/9) x = ±i√(7/9)

   • For u = 1:x^2 = 1 x = ±√1 x = ±1

   Reversing the substitution is a crucial step in the process, as it allows us to translate the solutions we found in terms of 'u' back into solutions for the original variable 'x'. This step highlights the cyclical nature of u-substitution - we transform the equation, solve it in the transformed space, and then reverse the transformation to find the solutions in the original space. The solutions for 'x' that arise from u = -7/9 involve the imaginary unit 'i', indicating that these solutions are complex numbers. This is a common occurrence when solving polynomial equations, and it underscores the importance of understanding the complex number system. The solutions for 'x' that arise from u = 1 are real numbers, representing the points where the original quartic function crosses the x-axis. The process of taking the square root introduces the ± sign, reflecting the fact that both positive and negative roots satisfy the equation. This is a subtle but crucial point in understanding the solutions to quadratic and higher-order equations. Each solution for 'x' represents a value that, when plugged back into the original equation, will make the equation true. These solutions are the fundamental building blocks of the equation's behavior, and they provide valuable insights into the nature of the mathematical relationship being described.

The Solutions and Their Significance

Therefore, the solutions to the equation 9x^4 - 2x^2 - 7 = 0 are x = ±i√(7/9) and x = ±1. Option B is correct. These solutions reveal the complete picture of the equation's behavior. The presence of both real and complex solutions is a characteristic feature of polynomial equations, and it reflects the richness and complexity of the mathematical world. The real solutions, x = ±1, represent the points where the graph of the quartic function intersects the x-axis. These are tangible, easily visualizable points on the real number line. The complex solutions, x = ±i√(7/9), on the other hand, exist in the realm of imaginary numbers. They do not have a direct representation on the real number line but are crucial for understanding the complete algebraic picture of the equation. Complex solutions often arise in systems that exhibit oscillatory or rotational behavior, and they play a vital role in fields such as electrical engineering and quantum mechanics. The combination of real and complex solutions provides a complete understanding of the equation's roots, which are the foundation for analyzing its behavior and applications. The solutions are not just numbers; they are key pieces of information that unlock the deeper meaning and significance of the equation.

Conclusion: Mastering u-Substitution

In conclusion, the solutions to the quartic equation 9x^4 - 2x^2 - 7 = 0 are x = ±i√(7/9) and x = ±1, obtained elegantly through the method of u-substitution. This technique is not merely a trick but a powerful tool in simplifying and solving polynomial equations. By recognizing the structure of the equation and applying the appropriate substitution, we transformed a seemingly complex problem into a manageable one. Mastering u-substitution equips us with a valuable problem-solving skill applicable to a wide range of mathematical contexts. It enhances our ability to tackle challenging equations and provides a deeper understanding of algebraic principles. The process of u-substitution highlights the beauty and elegance of mathematical problem-solving. It demonstrates how strategic manipulation and transformation can simplify complex problems and reveal their underlying structure. The ability to recognize patterns and apply appropriate techniques is a hallmark of mathematical proficiency. By mastering u-substitution, we not only gain the ability to solve specific types of equations but also develop a more general problem-solving mindset. This mindset is crucial for success in mathematics and other fields that require analytical thinking and logical reasoning. The journey of solving the quartic equation using u-substitution is a testament to the power of mathematical tools and the importance of developing a deep understanding of fundamental concepts. It underscores the fact that mathematics is not just about memorizing formulas but about developing a flexible and creative approach to problem-solving. This approach, honed through practice and experience, will serve us well in tackling future challenges and exploring the vast landscape of mathematical knowledge.