What Are The Solutions To The Equation X^4 - 9x^2 + 8 = 0, Solved Using U Substitution?

In this article, we'll delve into the process of finding the solutions to the quartic equation x^4 - 9x^2 + 8 = 0. Quartic equations, which are polynomial equations of the fourth degree, can sometimes seem daunting to solve. However, by employing a clever technique called u-substitution, we can transform this equation into a more manageable quadratic form. This method not only simplifies the equation but also provides a clear pathway to finding all its roots. We will walk through each step meticulously, ensuring a comprehensive understanding of the solution process. This detailed explanation is designed to assist students, educators, and anyone interested in enhancing their algebraic problem-solving skills. We will also discuss the underlying principles that make this substitution effective and explore why this approach is a valuable tool in solving higher-degree polynomial equations. By the end of this article, you will have a solid grasp of how to use substitution to solve quartic equations and a deeper appreciation for the elegance of algebraic manipulations. Understanding these techniques is crucial for mastering advanced mathematical concepts and tackling real-world problems that can be modeled using polynomial equations.

Understanding the U-Substitution Method

The u-substitution method is a powerful algebraic technique used to simplify complex equations by replacing a part of the expression with a single variable, typically u. This substitution transforms the original equation into a simpler form that is easier to solve. In the case of the quartic equation x^4 - 9x^2 + 8 = 0, we notice a pattern: the equation contains terms with x^4 and x^2. This suggests that we can substitute u = x^2, which will transform the equation into a quadratic form. The beauty of this method lies in its ability to reduce the complexity of higher-degree polynomial equations into quadratic equations, which we have well-established methods for solving. For instance, quadratic equations can be solved using factoring, completing the square, or the quadratic formula. This transformation not only simplifies the algebraic manipulation but also makes it easier to identify the roots of the equation. Moreover, understanding u-substitution provides a foundation for tackling more complex substitutions in calculus and other advanced mathematical fields. By mastering this technique, you can develop a strategic approach to problem-solving, recognizing patterns, and simplifying equations to find solutions more efficiently. This skill is particularly valuable in standardized tests and advanced coursework where time and accuracy are paramount.

Step 1: Substitute u = x^2

The first crucial step in solving the equation x^4 - 9x^2 + 8 = 0 using u-substitution is to recognize the inherent structure of the equation. We observe that the equation contains terms with x^4 and x^2, which are powers of x that differ by a factor of two. This observation is key because it allows us to make a strategic substitution. We let u = x^2. This substitution is the cornerstone of our approach because it directly transforms the quartic equation into a quadratic equation, which is a form we know how to handle. When we substitute u = x^2 into the original equation, we must also account for the x^4 term. Since u = x^2, it follows that u^2 = (x²⁾2 = x^4. Thus, we can replace x^4 with u^2 in our equation. This substitution effectively reduces the degree of the polynomial, making it more manageable. The underlying principle here is to simplify the equation by reducing its complexity, allowing us to apply familiar techniques. This step is not just about changing variables; it's about transforming the problem into a format that we can easily solve using existing methods. The ability to recognize such patterns and make appropriate substitutions is a fundamental skill in algebra and is essential for tackling more advanced mathematical problems. The substitution transforms the equation into a quadratic equation in terms of u, setting the stage for the next steps in our solution process.

Step 2: Rewrite the Equation in Terms of u

Following the substitution u = x^2, we now rewrite the original equation x^4 - 9x^2 + 8 = 0 in terms of u. As we established, x^4 becomes u^2, and x^2 becomes u. This transformation allows us to express the quartic equation as a quadratic equation in u. By replacing x^4 with u^2 and x^2 with u, the equation x^4 - 9x^2 + 8 = 0 is transformed into u^2 - 9u + 8 = 0. This quadratic equation is significantly easier to solve compared to the original quartic equation. The key advantage of this step is that we have reduced the complexity of the equation, making it amenable to standard quadratic solution techniques such as factoring, completing the square, or using the quadratic formula. Rewriting the equation in terms of u is not merely a symbolic manipulation; it is a strategic step that simplifies the problem and opens up a clear path to finding the solutions. This process highlights the power of algebraic transformations in simplifying complex problems. Understanding how to manipulate equations by changing variables is a crucial skill in mathematics. It allows us to tackle seemingly difficult problems by recasting them in a more familiar and solvable form. This skill is not only valuable in algebra but also in other areas of mathematics, such as calculus and differential equations, where substitution techniques are frequently used to simplify integrals and solve differential equations. With the equation now in quadratic form, we are well-prepared to find the values of u, which will then lead us to the solutions for x.

Step 3: Solve the Quadratic Equation for u

Now that we have the quadratic equation u^2 - 9u + 8 = 0, the next step is to solve for u. There are several methods to solve a quadratic equation, including factoring, completing the square, and using the quadratic formula. In this case, factoring is the most straightforward approach. We look for two numbers that multiply to 8 and add up to -9. These numbers are -1 and -8. Therefore, we can factor the quadratic equation as follows: (u - 1)(u - 8) = 0. This factored form of the equation provides us with a direct route to finding the values of u. According to the zero-product property, if the product of two factors is zero, then at least one of the factors must be zero. Thus, either u - 1 = 0 or u - 8 = 0. Solving these two linear equations gives us the values u = 1 and u = 8. These values of u are the solutions to the quadratic equation we obtained through substitution. However, it's crucial to remember that our original goal was to find the solutions for x. The values of u are intermediate results that we will use to find the values of x. This step demonstrates the power of factoring in solving quadratic equations. Factoring is an efficient method when the quadratic expression can be easily factored. However, if factoring is not immediately apparent, other methods such as the quadratic formula or completing the square can be used. The ability to choose the most efficient method for solving a quadratic equation is a valuable skill in algebra. With the values of u now determined, we are ready to reverse the substitution and find the corresponding values of x.

Step 4: Substitute Back to Find x

Having found the values of u, which are u = 1 and u = 8, we now need to substitute back to find the solutions for x. Recall that we initially made the substitution u = x^2. To find x, we need to solve the equations x^2 = 1 and x^2 = 8. For x^2 = 1, we take the square root of both sides, remembering to consider both the positive and negative roots. This gives us x = ±1. So, we have two solutions: x = 1 and x = -1. Next, we solve x^2 = 8 by taking the square root of both sides, again considering both positive and negative roots. This gives us x = ±√8. We can simplify √8 as √(4 * 2) = 2√2. Therefore, the solutions are x = ±2√2. So, we have two more solutions: x = 2√2 and x = -2√2. Combining all the solutions, we have x = 1, x = -1, x = 2√2, and x = -2√2. This step is crucial because it connects the intermediate solutions for u back to the original variable x, providing us with the complete set of solutions for the quartic equation. It also highlights the importance of considering both positive and negative roots when solving equations involving squares. The process of substituting back demonstrates the reversibility of the u-substitution method. We used substitution to simplify the equation and then reversed the substitution to find the solutions in terms of the original variable. This technique is a powerful tool in algebra and is applicable in many different contexts. With all the solutions for x determined, we have successfully solved the quartic equation using u-substitution.

Final Solutions

After completing the u-substitution and solving for x, we have found four solutions to the equation x^4 - 9x^2 + 8 = 0. These solutions are x = 1, x = -1, x = 2√2, and x = -2√2. It is important to note that a quartic equation, which is a polynomial equation of degree four, can have up to four solutions. In this case, we have found all four solutions, indicating that our solution process is complete and accurate. To summarize, we began by recognizing the structure of the quartic equation and making the substitution u = x^2. This transformed the equation into a quadratic equation in u, which we then solved by factoring. The solutions for u were u = 1 and u = 8. We then substituted back to find the solutions for x, obtaining x = ±1 and x = ±2√2. These are the four values of x that satisfy the original quartic equation. The solutions demonstrate the symmetry often found in polynomial equations, with both positive and negative values being solutions. This symmetry arises from the even powers of x in the equation. In conclusion, the solutions to the equation x^4 - 9x^2 + 8 = 0 are x = 1, x = -1, x = 2√2, and x = -2√2, which corresponds to option B. This process illustrates the effectiveness of the u-substitution method in simplifying and solving higher-degree polynomial equations.

Therefore, the correct answer is B: x = ±1 and x = ±2√2