Equivalent Quadratic Equation With Substitution

Let's explore how to simplify a given quadratic equation using substitution. Specifically, we will focus on the equation $(x-4)^2 - (x-4) - 6 = 0$ and determine which of the provided options correctly represents the equation after substituting $u = (x-4)$. This involves understanding the basics of algebraic substitution and how it can be used to transform equations into simpler forms.

Understanding the Substitution

The heart of this problem lies in the technique of substitution. Substitution is a powerful tool in algebra that allows us to replace a complex expression with a single variable, making the equation easier to manipulate and solve. In our case, we are given the substitution $u = (x-4)$. This means wherever we see $(x-4)$ in the original equation, we can replace it with $u$. This simplifies the equation and can make it easier to recognize its structure and solve for the variable.

When we apply the substitution, we aim to transform the original equation into a new equation that is expressed in terms of $u$. This new equation should be equivalent to the original, meaning that it has the same solutions, just expressed in terms of a different variable. The goal is to correctly perform the substitution and simplify the resulting equation without altering its fundamental properties.

By carefully substituting and simplifying, we can identify the correct equivalent quadratic equation from the given options. This process highlights the importance of paying close attention to detail and understanding the underlying algebraic principles.

Analyzing the Original Equation

Our starting point is the equation $(x-4)^2 - (x-4) - 6 = 0$. Before we dive into the substitution, let's take a closer look at this equation. We can see that the expression $(x-4)$ appears twice. This repetition is a key indicator that substitution might be a useful strategy. The equation is a quadratic in the sense that it involves a squared term, namely $(x-4)^2$. However, it's not in the standard quadratic form $ax^2 + bx + c = 0$ directly, but rather a variation of it. Recognizing this structure helps us to appreciate how the substitution will simplify things.

Breaking down the equation further, we have three terms: $(x-4)^2$, $-(x-4)$, and $-6$. The first term is the square of $(x-4)$, the second term is the negative of $(x-4)$, and the third term is a constant. These terms combine to form the quadratic expression that we are working with.

Understanding the components of the original equation allows us to correctly apply the substitution. It's crucial to keep track of each term and ensure that the substitution is applied consistently throughout the equation. This careful approach will help us avoid errors and arrive at the correct equivalent equation.

Performing the Substitution

Now, let's perform the substitution $u = (x-4)$ in the original equation $(x-4)^2 - (x-4) - 6 = 0$. Wherever we encounter $(x-4)$, we replace it with $u$. This gives us:

$u^2 - u - 6 = 0$

This resulting equation is a quadratic equation in terms of $u$. It's significantly simpler than the original equation, making it easier to analyze and solve if needed. This demonstrates the power of substitution in simplifying complex algebraic expressions.

The substitution transforms the original equation by replacing the expression $(x-4)$ with a single variable $u$. This process maintains the equivalence of the equation while making it more manageable. The key is to correctly identify the expression to be substituted and apply the substitution consistently throughout the equation.

The resulting equation, $u^2 - u - 6 = 0$, is a direct result of the substitution. It is a quadratic equation in standard form, which makes it easier to work with. This highlights the effectiveness of substitution as a problem-solving technique in algebra.

Analyzing the Answer Choices

Now, let's compare our result with the given answer choices:

A. $u^2 - (u-4) - 6 = 0$ where $u = (x-4)$ B. $u^2 - 16 - u - 6 = 0$ where $u = (x-4)$ C. $(u-4)^2 - (u-4) - 6 = 0$ where $u = (x-4)$ D. $u^2 - u - 6 = 0$ where $u = (x-4)$

By comparing our result $u^2 - u - 6 = 0$ with the answer choices, we can see that option D matches our result exactly. The other options have either incorrectly substituted or introduced additional terms that are not present in the correct substitution.

Option A incorrectly retains a $(u-4)$ term, which is not a direct substitution. Option B introduces a $-16$ term, suggesting an incorrect expansion or substitution. Option C substitutes $u-4$ instead of $u$ which is incorrect based on the given $u = (x-4)$. Therefore, options A, B, and C are incorrect.

Our analysis confirms that option D, $u^2 - u - 6 = 0$, is the correct equivalent equation after performing the substitution $u = (x-4)$. This careful comparison of the answer choices with our result is crucial for arriving at the correct answer.

Conclusion

In conclusion, the quadratic equation equivalent to $(x-4)^2 - (x-4) - 6 = 0$ after substituting $u = (x-4)$ is $u^2 - u - 6 = 0$. Therefore, the correct answer is option D.

This problem illustrates the power and utility of algebraic substitution in simplifying equations. By correctly identifying the expression to be substituted and applying the substitution consistently, we can transform complex equations into more manageable forms. This technique is a valuable tool in algebra and can be applied to a wide range of problems.

Understanding the basics of algebraic manipulation, such as substitution, is essential for solving various mathematical problems. The ability to simplify equations and express them in different forms is a fundamental skill that can greatly enhance your problem-solving abilities. This example demonstrates the importance of paying attention to detail and applying the correct algebraic principles to arrive at the correct solution.

To further enhance your understanding of quadratic equations and algebraic substitutions, consider exploring resources like Khan Academy's section on quadratic equations: Khan Academy Quadratic Equations.