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

This article delves into the step-by-step solution of the equation $(x+2)^2 + 12(x+2) - 14 = 0$. We will employ a strategic approach using u-substitution to simplify the equation and then utilize the quadratic formula to find the roots. Understanding these techniques is crucial for mastering algebraic manipulations and solving quadratic equations effectively. This comprehensive guide aims to provide a clear and concise explanation of each step, ensuring a solid understanding of the solution process. Whether you're a student tackling algebra problems or simply looking to refresh your mathematical skills, this article offers valuable insights and practical methods for solving complex equations.

1. Introduction to the Problem

We are presented with the equation $(x+2)^2 + 12(x+2) - 14 = 0$. This equation, at first glance, appears to be a complex quadratic form. However, by recognizing the repeating term $(x+2)$, we can simplify it using a technique called u-substitution. This method allows us to transform the equation into a more manageable quadratic equation in terms of a new variable, 'u'. This initial simplification is key to unlocking the solution. The goal is to find the values of 'x' that satisfy the original equation. By mastering this type of problem, you'll enhance your ability to solve a variety of algebraic equations, making this a valuable skill in mathematics. The ability to identify patterns and make strategic substitutions is a cornerstone of effective problem-solving in algebra and beyond.

2. Applying u-Substitution

The first key step in solving this equation is to simplify it using u-substitution. We let $u = (x+2)$. This substitution transforms the original equation into a much simpler quadratic form. Substituting 'u' into the equation, we get: $u^2 + 12u - 14 = 0$. This new equation is a standard quadratic equation in the variable 'u', which is much easier to work with than the original. The power of u-substitution lies in its ability to convert complex expressions into more familiar forms, making them accessible to standard solution methods. By replacing the repetitive term $(x+2)$ with 'u', we've effectively reduced the complexity of the problem. This technique is widely used in algebra and calculus to simplify equations and integrals, making it a fundamental tool in mathematical problem-solving. The transformation allows us to focus on solving a standard quadratic equation before reverting back to the original variable 'x'.

3. Using the Quadratic Formula

Now that we have the quadratic equation $u^2 + 12u - 14 = 0$, we can solve for 'u' using the quadratic formula. The quadratic formula is a universal tool for finding the roots of any quadratic equation in the form $ax^2 + bx + c = 0$, and it's given by: $x = \frac{-b ${}$ \sqrt{b^2 - 4ac}}{2a}$. In our case, the coefficients are: a = 1, b = 12, and c = -14. Substituting these values into the quadratic formula, we get:

$u = \frac{-12 \pm \sqrt{12^2 - 4(1)(-14)}}{2(1)}$

Simplifying the expression under the square root:

$u = \frac{-12 \pm \sqrt{144 + 56}}{2}$

$u = \frac{-12 \pm \sqrt{200}}{2}$

Further simplifying the square root:

$u = \frac{-12 \pm \sqrt{100 * 2}}{2}$

$u = \frac{-12 \pm 10\sqrt{2}}{2}$

Finally, dividing both terms in the numerator by 2:

$u = -6 \pm 5\sqrt{2}$

Therefore, we have two possible values for 'u': $u = -6 + 5\sqrt{2}$ and $u = -6 - 5\sqrt{2}$. The quadratic formula is a powerful method, ensuring that we can find the solutions to any quadratic equation, regardless of its complexity. This step is crucial in bridging the gap between the simplified equation in terms of 'u' and the ultimate solution for 'x'.

4. Substituting Back to Find x

Having found the values of 'u', we now need to substitute back to find the values of 'x'. Recall that we defined $u = (x+2)$. So, we have two equations to solve:

1. $x + 2 = -6 + 5\sqrt{2}$
2. $x + 2 = -6 - 5\sqrt{2}$

Solving the first equation for 'x':

$x = -6 + 5\sqrt{2} - 2$

$x = -8 + 5\sqrt{2}$

Solving the second equation for 'x':

$x = -6 - 5\sqrt{2} - 2$

$x = -8 - 5\sqrt{2}$

Thus, the two solutions for 'x' are $x = -8 + 5\sqrt{2}$ and $x = -8 - 5\sqrt{2}$. This reverse substitution is a critical step in the u-substitution method. It allows us to transition from the temporary variable 'u' back to the original variable 'x', providing the final solution to the initial equation. Without this step, we would only have solutions for 'u', which do not answer the original question. The ability to perform this substitution correctly demonstrates a thorough understanding of the u-substitution technique.

5. The Solutions

Therefore, the solutions to the equation $(x+2)^2 + 12(x+2) - 14 = 0$ are $x = -8 + 5\sqrt{2}$ and $x = -8 - 5\sqrt{2}$. This can be written in a more compact form as $x = -8 \pm 5\sqrt{2}$. This means the correct answer among the given options is A. $x=-8 \pm 5 \sqrt{2}$. Presenting the solutions in a clear and concise manner is essential for effective communication of mathematical results. We have successfully used u-substitution to transform a complex equation into a simpler quadratic form, solved for the temporary variable, and then substituted back to find the solutions for the original variable. This process highlights the power of algebraic manipulation techniques in simplifying and solving seemingly difficult problems. The ability to arrive at the correct solutions demonstrates a strong grasp of both the u-substitution method and the quadratic formula.

6. Conclusion

In conclusion, we have successfully solved the equation $(x+2)^2 + 12(x+2) - 14 = 0$ by employing the techniques of u-substitution and the quadratic formula. This problem demonstrates a powerful approach to handling complex algebraic expressions by simplifying them into more manageable forms. The u-substitution allowed us to transform the original equation into a standard quadratic equation, which we then solved using the quadratic formula. Finally, substituting back, we found the solutions for 'x'. Mastering these techniques is fundamental for tackling various algebraic problems, showcasing the importance of strategic problem-solving in mathematics. This step-by-step solution provides a comprehensive understanding of the process, enabling readers to apply these methods to similar problems confidently. The combination of algebraic manipulation and formula application is a cornerstone of mathematical proficiency.