Solving X^4 + 3x^2 + 2 = 0 With U-Substitution A Step-by-Step Guide

Jul 12, 2025 by ADMIN 68 views

Solving the Equation x^4 + 3x^2 + 2 = 0 Using u-Substitution

In the realm of algebra, solving polynomial equations is a fundamental skill. Among these, quartic equations, those with a highest degree of four, can present a unique challenge. However, with strategic techniques like u-substitution, we can transform seemingly complex equations into more manageable forms. This article delves into the process of solving the quartic equation x^4 + 3x^2 + 2 = 0 using u-substitution, providing a step-by-step guide to arrive at the solutions.

Understanding the Equation

Before diving into the solution, let's first understand the equation we're dealing with. The equation x^4 + 3x^2 + 2 = 0 is a quartic equation because the highest power of the variable x is 4. Notice also that this equation has a special form: it's a quadratic in x². This observation is key to our solution strategy.

Identifying the Quadratic Form

The equation can be rewritten to highlight its quadratic nature. We can express x⁴ as (x²)² , making the equation look like this: (x²)² + 3(x²) + 2 = 0. This form closely resembles a standard quadratic equation, which we can exploit using u-substitution.

The Power of u-Substitution

U-substitution is a technique used to simplify equations by replacing a complex expression with a single variable, in this case, u. This substitution transforms the equation into a more familiar form, making it easier to solve. For our equation, the natural substitution is to let u = x². This will transform our quartic equation into a quadratic equation in terms of u.

Step-by-Step Solution

Now, let's walk through the process of solving the equation x^4 + 3x^2 + 2 = 0 using u-substitution.

1. Apply the u-Substitution

As mentioned earlier, we'll substitute u = x². Replacing every instance of x² with u in the equation gives us: u² + 3u + 2 = 0. This is now a standard quadratic equation that we can easily solve.

2. Solve the Quadratic Equation in u

We can solve the quadratic equation u² + 3u + 2 = 0 using several methods, such as factoring, completing the square, or the quadratic formula. In this case, factoring is the most straightforward approach. We need to find two numbers that multiply to 2 and add up to 3. These numbers are 1 and 2. Therefore, we can factor the quadratic equation as: (u + 1)(u + 2) = 0. Setting each factor equal to zero gives us the solutions for u: u + 1 = 0 => u = -1, u + 2 = 0 => u = -2. So, the solutions for u are u = -1 and u = -2.

3. Substitute Back to Find x

Remember that we introduced u as a substitution for x². Now, we need to substitute back to find the values of x. We have two equations to solve: x² = -1 and x² = -2. For x² = -1, taking the square root of both sides gives us x = ±√(-1) = ±i, where i is the imaginary unit (√(-1)). For x² = -2, taking the square root of both sides gives us x = ±√(-2) = ±i√2. Therefore, the solutions for x are x = ±i and x = ±i√2.

Verifying the Solutions

It's always a good practice to verify the solutions by plugging them back into the original equation. Let's verify our solutions: x = i, x = -i, x = i√2, and x = -i√2.

Verification for x = i

Substituting x = i into the original equation: (i)⁴ + 3(i)² + 2 = 1 + 3(-1) + 2 = 1 - 3 + 2 = 0. This confirms that x = i is a solution.

Verification for x = -i

Substituting x = -i into the original equation: (-i)⁴ + 3(-i)² + 2 = 1 + 3(-1) + 2 = 1 - 3 + 2 = 0. This confirms that x = -i is a solution.

Verification for x = i√2

Substituting x = i√2 into the original equation: (i√2)⁴ + 3(i√2)² + 2 = (4) + 3(-2) + 2 = 4 - 6 + 2 = 0. This confirms that x = i√2 is a solution.

Verification for x = -i√2

Substituting x = -i√2 into the original equation: (-i√2)⁴ + 3(-i√2)² + 2 = (4) + 3(-2) + 2 = 4 - 6 + 2 = 0. This confirms that x = -i√2 is a solution.

Conclusion

By employing the technique of u-substitution, we successfully transformed the quartic equation x^4 + 3x^2 + 2 = 0 into a solvable quadratic equation. This allowed us to find the four solutions: x = i, x = -i, x = i√2, and x = -i√2. These solutions are complex numbers, which is not uncommon for quartic equations. The process highlights the power of algebraic manipulation and the importance of recognizing patterns in equations. U-substitution is a valuable tool in your algebraic arsenal, allowing you to tackle a wide range of equations with greater confidence. Understanding and applying these techniques is crucial for success in higher-level mathematics and related fields.

Understanding how to solve polynomial equations is a cornerstone of algebra. This article will provide an in-depth look at solving the quartic equation x⁴ + 3x² + 2 = 0. We will focus on the method of u-substitution, which simplifies the equation and makes it easier to solve. Quartic equations, those with a degree of four, can seem daunting at first, but with the right techniques, they can be tackled systematically. This guide is designed to provide a clear and comprehensive explanation of the process, ensuring you grasp the underlying concepts and can apply them to similar problems.

Preliminary Analysis of the Quartic Equation

Before we jump into the solution, let's examine the equation x⁴ + 3x² + 2 = 0 more closely. The first thing to notice is that while it is a quartic equation, it has a specific structure that makes it amenable to a particular solution method. Specifically, this equation is a quadratic in x². This means that the powers of x appear in even degrees (4 and 2), and there is a constant term. This structure allows us to use a substitution technique to reduce the equation to a simpler form. Recognizing these patterns is an essential first step in problem-solving in mathematics. It allows us to choose the most efficient and effective method to find the solutions.

The Strategic Importance of U-Substitution

U-substitution is a powerful technique used to simplify equations by replacing a complex expression with a single variable, usually u. The goal is to transform the equation into a more recognizable form, such as a quadratic, which we can then solve using standard methods. In our case, the strategic choice is to substitute u = x². This substitution will transform the quartic equation into a quadratic equation in terms of u. This approach is a prime example of how algebraic manipulation can make a complex problem much more manageable. The key is to identify the appropriate substitution that will simplify the equation while preserving its fundamental structure.

Transforming the Equation with U-Substitution

By substituting u = x² into the equation x⁴ + 3x² + 2 = 0, we replace x⁴ with u² and x² with u. This transformation gives us the new equation: u² + 3u + 2 = 0. This equation is a standard quadratic equation, which we can solve using various methods, such as factoring, completing the square, or the quadratic formula. The simplicity of this quadratic equation compared to the original quartic equation demonstrates the effectiveness of the u-substitution technique. It allows us to break down a complex problem into simpler, more manageable steps.

Solving the Transformed Quadratic Equation

Now that we have the quadratic equation u² + 3u + 2 = 0, we can proceed to solve it for u. There are several methods we could use, but factoring is the most efficient in this case. Factoring involves finding two numbers that multiply to the constant term (2) and add up to the coefficient of the linear term (3). These numbers are 1 and 2. Therefore, we can factor the quadratic equation as: (u + 1)(u + 2) = 0. This factorization allows us to easily find the solutions for u by setting each factor equal to zero.

Finding the Solutions for u

Setting each factor in the equation (u + 1)(u + 2) = 0 equal to zero gives us two simple linear equations: u + 1 = 0 and u + 2 = 0. Solving these equations for u gives us the solutions: u = -1 and u = -2. These are the values of u that satisfy the quadratic equation. However, our original goal was to find the solutions for x, so we must now substitute back to find the corresponding values of x.

The Critical Step of Substituting Back

Recall that we made the substitution u = x². Now, we need to reverse this process to find the values of x. We have two equations to solve: x² = -1 and x² = -2. These equations tell us that the square of x is a negative number, which means that the solutions for x will be complex numbers. This is a common occurrence when solving polynomial equations of degree greater than 2. The solutions may be real, complex, or a combination of both. In our case, we will find complex solutions.

Determining the Solutions for x

To find the solutions for x, we need to take the square root of both sides of the equations x² = -1 and x² = -2. For x² = -1, taking the square root gives us x = ±√(-1) = ±i, where i is the imaginary unit (√(-1)). For x² = -2, taking the square root gives us x = ±√(-2) = ±i√2. Therefore, the solutions for x are x = ±i and x = ±i√2. These are the four solutions to the original quartic equation. They are all complex numbers, which is consistent with the fact that the quadratic equation in u had negative solutions.

Summarizing the Complex Solutions

The solutions to the equation x⁴ + 3x² + 2 = 0 are x = i, x = -i, x = i√2, and x = -i√2. These are complex numbers, each consisting of an imaginary part. Complex solutions often arise in polynomial equations, especially those of higher degrees. It is important to understand how to work with complex numbers to solve these equations correctly. The u-substitution method allowed us to transform a complex quartic equation into a simpler quadratic equation, making the solution process much more straightforward.

Verification of the Solutions

It's essential to verify our solutions to ensure they are correct. We can do this by substituting each solution back into the original equation and checking if it holds true. This process helps to catch any errors that may have occurred during the solution process. Verification is a crucial step in problem-solving, as it provides confidence in the accuracy of the results.

Verifying the Solutions Step-by-Step

Let's verify each of our solutions: x = i, x = -i, x = i√2, and x = -i√2.

Verification for x = i

Substituting x = i into the original equation: (i)⁴ + 3(i)² + 2 = 1 + 3(-1) + 2 = 1 - 3 + 2 = 0. This confirms that x = i is a solution.

Verification for x = -i

Substituting x = -i into the original equation: (-i)⁴ + 3(-i)² + 2 = 1 + 3(-1) + 2 = 1 - 3 + 2 = 0. This confirms that x = -i is a solution.

Verification for x = i√2

Substituting x = i√2 into the original equation: (i√2)⁴ + 3(i√2)² + 2 = (4) + 3(-2) + 2 = 4 - 6 + 2 = 0. This confirms that x = i√2 is a solution.

Verification for x = -i√2

Substituting x = -i√2 into the original equation: (-i√2)⁴ + 3(-i√2)² + 2 = (4) + 3(-2) + 2 = 4 - 6 + 2 = 0. This confirms that x = -i√2 is a solution.

The Importance of Verification

Our verification process has confirmed that all four solutions satisfy the original equation. This gives us confidence in the correctness of our solution. Verification is a crucial step in mathematical problem-solving. It ensures that the obtained solutions are accurate and valid. It is particularly important when dealing with complex equations or multiple steps, where errors are more likely to occur. By taking the time to verify our solutions, we can be confident in our results.

Final Conclusion on Solving the Quartic Equation

In conclusion, we have successfully solved the quartic equation x⁴ + 3x² + 2 = 0 using the u-substitution technique. This method allowed us to transform the equation into a simpler quadratic form, which we could then solve using standard methods. We found the four solutions to be x = i, x = -i, x = i√2, and x = -i√2, all of which are complex numbers. We also verified these solutions by substituting them back into the original equation. This comprehensive guide has demonstrated the power of algebraic manipulation and the importance of choosing the right techniques to solve mathematical problems. The u-substitution method is a valuable tool in algebra, allowing us to tackle complex equations with greater efficiency and accuracy.

Solving equations is a fundamental skill in mathematics, and polynomial equations are a significant class of equations to understand. Quartic equations, which have a highest degree of four, can sometimes appear challenging, but with the right techniques, they become manageable. This article will focus on solving the equation x^4 + 3x^2 + 2 = 0 using the method of u-substitution. This method is particularly useful for equations that can be transformed into a quadratic form. By breaking down the process step-by-step, this guide will provide a clear understanding of how to approach such equations and find their solutions.

Recognizing the Structure of the Equation

The equation x^4 + 3x^2 + 2 = 0 is a quartic equation. However, it has a special form that allows us to use a simplification technique. Notice that the equation only contains even powers of x: x⁴ and x². This structure suggests that we can rewrite the equation in terms of x². Specifically, we can express x⁴ as (x²)². This transformation highlights the quadratic nature of the equation. Recognizing this structure is crucial because it leads us to the most efficient solution method. Many mathematical problems can be simplified by carefully observing their structure and choosing the appropriate techniques.

The Strategic Use of U-Substitution Explained

U-substitution is a technique that simplifies equations by replacing a complex expression with a single variable, often denoted by u. The goal is to transform the equation into a more familiar form, such as a quadratic equation, which we know how to solve. In the case of our equation, the appropriate substitution is u = x². This substitution will replace x² with u and (x²)² with u², transforming the quartic equation into a quadratic equation in u. This strategic substitution is a powerful tool in algebra, allowing us to solve a wide range of equations. The key is to identify the correct substitution that simplifies the equation while preserving its essential properties.

Transforming the Quartic into a Quadratic

By substituting u = x² into the equation x^4 + 3x^2 + 2 = 0, we obtain the new equation: u² + 3u + 2 = 0. This equation is a standard quadratic equation, which we can solve using various methods such as factoring, completing the square, or the quadratic formula. The transformation from a quartic to a quadratic equation demonstrates the effectiveness of the u-substitution technique. It allows us to break down a complex problem into simpler, more manageable steps. This approach is a fundamental strategy in mathematics: reducing a complex problem to a simpler one that we already know how to solve.

Solving the Quadratic Equation in Terms of u

Now that we have the quadratic equation u² + 3u + 2 = 0, we can proceed to find its solutions. Factoring is the most efficient method for this particular equation. Factoring involves finding two numbers that multiply to the constant term (2) and add up to the coefficient of the linear term (3). These numbers are 1 and 2. Therefore, we can factor the quadratic equation as: (u + 1)(u + 2) = 0. This factorization makes it easy to find the solutions for u by setting each factor equal to zero.

Finding the Values of u

Setting each factor in the equation (u + 1)(u + 2) = 0 equal to zero gives us two simple linear equations: u + 1 = 0 and u + 2 = 0. Solving these equations for u gives us the solutions: u = -1 and u = -2. These are the values of u that satisfy the quadratic equation. However, our ultimate goal is to find the values of x, so we must now reverse the substitution process to find the corresponding values of x.

Substituting Back to Find the Solutions for x

Remember that we made the substitution u = x². Now, we need to reverse this substitution to find the values of x. We have two equations to solve: x² = -1 and x² = -2. These equations tell us that the square of x is a negative number, which means that the solutions for x will be complex numbers. This is a common outcome when solving polynomial equations, especially those with even degrees. Complex solutions indicate that the equation has no real roots, but it does have solutions in the complex number system. Understanding complex numbers is essential for solving many algebraic equations.

Determining the Complex Solutions for x

Summarizing the Complex Solutions for x

The solutions to the equation x^4 + 3x^2 + 2 = 0 are x = i, x = -i, x = i√2, and x = -i√2. These complex solutions highlight the importance of understanding the complex number system in algebra. Complex numbers are numbers of the form a + bi, where a and b are real numbers and i is the imaginary unit (√(-1)). They are essential for solving many polynomial equations, especially those that do not have real roots. The u-substitution method allowed us to transform a seemingly complex quartic equation into a more manageable quadratic equation, making the solution process much clearer.

The Importance of Verifying the Solutions

It is crucial to verify our solutions to ensure that they are correct. Verification involves substituting each solution back into the original equation and checking if it holds true. This process helps to identify any errors that may have occurred during the solution process. Verification is a fundamental step in problem-solving, as it provides confidence in the accuracy of the results. It is particularly important when dealing with complex equations or when multiple steps are involved in the solution.

The Step-by-Step Process of Verification

Let's verify each of our solutions: x = i, x = -i, x = i√2, and x = -i√2.

Verification for x = i

Substituting x = i into the original equation: (i)⁴ + 3(i)² + 2 = 1 + 3(-1) + 2 = 1 - 3 + 2 = 0. This confirms that x = i is a solution.

Verification for x = -i

Substituting x = -i into the original equation: (-i)⁴ + 3(-i)² + 2 = 1 + 3(-1) + 2 = 1 - 3 + 2 = 0. This confirms that x = -i is a solution.

Verification for x = i√2

Substituting x = i√2 into the original equation: (i√2)⁴ + 3(i√2)² + 2 = (4) + 3(-2) + 2 = 4 - 6 + 2 = 0. This confirms that x = i√2 is a solution.

Verification for x = -i√2

Substituting x = -i√2 into the original equation: (-i√2)⁴ + 3(-i√2)² + 2 = (4) + 3(-2) + 2 = 4 - 6 + 2 = 0. This confirms that x = -i√2 is a solution.

Why Verification Matters in Mathematics

Our verification process has demonstrated that all four solutions satisfy the original equation. This gives us confidence in the correctness of our solution. Verification is a crucial step in mathematical problem-solving. It helps ensure that the obtained solutions are accurate and valid. It is especially important when dealing with complex equations or multiple steps, where errors are more likely to occur. By taking the time to verify our solutions, we can be confident in our results and avoid making mistakes.

Final Thoughts on Solving the Equation

In conclusion, we have successfully solved the quartic equation x^4 + 3x^2 + 2 = 0 using the u-substitution technique. This method allowed us to transform the equation into a simpler quadratic form, which we could then solve using standard methods. We found the four solutions to be x = i, x = -i, x = i√2, and x = -i√2, all of which are complex numbers. We also verified these solutions by substituting them back into the original equation. This guide has provided a detailed explanation of the process, highlighting the importance of recognizing the structure of the equation, choosing the appropriate solution method, and verifying the results. The u-substitution method is a valuable tool in algebra, allowing us to tackle complex equations with greater efficiency and accuracy. By mastering this technique, you can confidently solve a wide range of polynomial equations.