Solving (x+2)^2 + 12(x+2) - 14 = 0 With U-Substitution And Quadratic Formula

This article provides a comprehensive solution to the equation (x+2)^2 + 12(x+2) - 14 = 0. We will employ u-substitution to simplify the equation and then utilize the quadratic formula to find the solutions for x. This method allows us to transform a seemingly complex equation into a manageable quadratic form, making it easier to solve.

Introduction to u-Substitution

The u-substitution technique is a powerful tool in algebra that simplifies equations by replacing a complex expression with a single variable, often 'u'. This substitution transforms the equation into a more recognizable form, such as a quadratic equation, which we can then solve using standard methods. In this case, we observe that the expression (x+2) appears twice in the equation, making u-substitution an ideal approach. By letting u = (x+2), we can rewrite the given equation in terms of 'u', resulting in a simpler quadratic equation that is easier to manipulate and solve. This initial step is crucial as it lays the foundation for applying the quadratic formula later on. The beauty of u-substitution lies in its ability to break down complex problems into smaller, more manageable parts, allowing us to focus on solving the core equation without being overwhelmed by the original complexity. This technique is not only applicable to quadratic equations but also extends to various other types of equations in calculus and other advanced mathematical fields. Mastering u-substitution is a valuable skill for any student or professional dealing with mathematical problem-solving, providing a systematic way to approach and simplify challenging expressions.

Applying u-Substitution

To begin solving the equation (x+2)^2 + 12(x+2) - 14 = 0, we will apply the u-substitution method. Let's set u = (x+2). This substitution allows us to rewrite the equation in terms of u, making it a simpler quadratic equation. Replacing every instance of (x+2) with u, the equation transforms into: u^2 + 12u - 14 = 0. This resulting equation is a standard quadratic equation in the form of au^2 + bu + c = 0, where a = 1, b = 12, and c = -14. This transformation is a crucial step as it simplifies the original equation, making it more amenable to solving using the quadratic formula. By performing u-substitution, we've effectively converted a complex expression into a more manageable form, allowing us to apply well-established techniques for solving quadratic equations. The simplicity of the resulting quadratic equation makes the subsequent steps, such as applying the quadratic formula, much easier and less prone to errors. Understanding and applying u-substitution is a fundamental skill in algebra, enabling us to tackle more complex problems by breaking them down into simpler, solvable components. This technique is not only useful for solving equations but also in various other mathematical contexts, such as integration in calculus.

Using the Quadratic Formula

Now that we have the quadratic equation in terms of u, which is u^2 + 12u - 14 = 0, we can apply the quadratic formula to solve for u. The quadratic formula is a fundamental tool for finding the roots of any quadratic equation in the form au^2 + bu + c = 0. The formula is given by: u = (-b ± √(b^2 - 4ac)) / (2a). In our equation, a = 1, b = 12, and c = -14. Substituting these values into the formula, we get: u = (-12 ± √(12^2 - 4 * 1 * -14)) / (2 * 1). Simplifying the expression under the square root: 12^2 - 4 * 1 * -14 = 144 + 56 = 200. Thus, the equation becomes: u = (-12 ± √200) / 2. Further simplifying the square root, we recognize that 200 can be expressed as 100 * 2, and the square root of 100 is 10. So, √200 = √(100 * 2) = 10√2. Substituting this back into the equation, we have: u = (-12 ± 10√2) / 2. Finally, we can divide both terms in the numerator by 2 to get: u = -6 ± 5√2. Therefore, the solutions for u are u = -6 + 5√2 and u = -6 - 5√2. This step is critical in the process of solving the original equation as it provides us with the values of u that satisfy the transformed quadratic equation. The quadratic formula is a cornerstone of algebra, providing a reliable method for solving quadratic equations regardless of their complexity. Its application here demonstrates its power and versatility in handling mathematical problems.

Substituting Back to Solve for x

Having found the values of u, we now need to substitute back to find the values of x. Recall that we initially defined u = (x+2). We have two solutions for u: u = -6 + 5√2 and u = -6 - 5√2. For each value of u, we will solve for x. First, let's consider u = -6 + 5√2. Substituting this into our equation u = (x+2), we get: -6 + 5√2 = x + 2. Solving for x, we subtract 2 from both sides: x = -6 + 5√2 - 2, which simplifies to x = -8 + 5√2. Next, we consider the second value, u = -6 - 5√2. Substituting this into u = (x+2), we get: -6 - 5√2 = x + 2. Again, solving for x by subtracting 2 from both sides: x = -6 - 5√2 - 2, which simplifies to x = -8 - 5√2. Therefore, the two solutions for x are x = -8 + 5√2 and x = -8 - 5√2. These solutions represent the values of x that satisfy the original equation (x+2)^2 + 12(x+2) - 14 = 0. The process of substituting back is a crucial step in u-substitution, as it allows us to express the solutions in terms of the original variable, providing a complete solution to the problem. This step underscores the effectiveness of u-substitution as a technique for simplifying complex equations and finding their solutions.

Final Solutions and Conclusion

In conclusion, the solutions to the equation (x+2)^2 + 12(x+2) - 14 = 0 are x = -8 + 5√2 and x = -8 - 5√2. This can be written more concisely as x = -8 ± 5√2. We arrived at these solutions by employing u-substitution to simplify the equation and then utilizing the quadratic formula. The u-substitution technique allowed us to transform the original equation into a standard quadratic form, making it easier to solve. By letting u = (x+2), we converted the equation into u^2 + 12u - 14 = 0. We then applied the quadratic formula to find the values of u, which were u = -6 ± 5√2. Finally, we substituted back to find the values of x, yielding x = -8 ± 5√2. This step-by-step process demonstrates the power and elegance of algebraic techniques in solving complex equations. The combination of u-substitution and the quadratic formula provides a robust method for tackling a wide range of mathematical problems. Understanding and mastering these techniques is essential for anyone pursuing further studies in mathematics or related fields. The solutions we have found are precise and represent the exact values of x that satisfy the given equation, highlighting the importance of careful and methodical problem-solving in mathematics. The ability to break down complex problems into smaller, more manageable steps is a key skill in mathematical problem-solving, and this example effectively illustrates that approach.

Therefore, the correct answer is:

A. x = -8 ± 5√2