Solving Algebraic Equations Finding A If (x^2-6x+10)/(x-2)=A+2/(x-2)

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This article delves into the solution of an intriguing algebraic equation: If (x^2-6x+10)/(x-2)=A+2/(x-2) for all values of x ≠ 2, which of these is equal to A? This problem, often encountered in mathematics, tests our understanding of algebraic manipulation and polynomial division. We will explore the step-by-step process of solving this equation, ensuring a comprehensive understanding for readers of all backgrounds. Let's embark on this mathematical journey together!

The core of our discussion lies in the equation: (x^2-6x+10)/(x-2)=A+2/(x-2). The challenge is to find an expression equivalent to A. This equation holds true for all values of x except x = 2, as the denominator would become zero, rendering the expression undefined. Our mission is to isolate A and determine its algebraic form. The multiple-choice options provided are:

(A) x + 1 (B) x + 4 (C) x - 1 (D) x - 4

To solve this, we will employ algebraic techniques to simplify the equation and identify the correct expression for A.

Our primary approach involves manipulating the given equation to isolate A. We begin with the equation:

(x^2 - 6x + 10) / (x - 2) = A + 2 / (x - 2)

To isolate A, we subtract 2 / (x - 2) from both sides of the equation:

(x^2 - 6x + 10) / (x - 2) - 2 / (x - 2) = A

Now, we combine the fractions on the left side, since they share a common denominator:

[(x^2 - 6x + 10) - 2] / (x - 2) = A

Simplifying the numerator, we get:

(x^2 - 6x + 8) / (x - 2) = A

Next, we factor the quadratic expression in the numerator. We are looking for two numbers that multiply to 8 and add up to -6. These numbers are -4 and -2. Therefore, we can factor the numerator as:

[(x - 4)(x - 2)] / (x - 2) = A

Now, we can cancel the common factor of (x - 2) from the numerator and the denominator, keeping in mind that x ≠ 2:

x - 4 = A

Thus, we find that A is equal to x - 4. This corresponds to option (D).

This method showcases the power of algebraic manipulation in simplifying complex equations. By carefully applying the rules of algebra, we were able to isolate A and determine its equivalent expression. This approach not only provides the correct answer but also reinforces fundamental algebraic skills, crucial for tackling a wide range of mathematical problems.

An alternative approach to solving this problem involves polynomial long division. We can divide the polynomial x² - 6x + 10 by x - 2 to find the quotient and remainder. This method provides a structured way to simplify rational expressions and can be particularly useful when dealing with higher-degree polynomials.

Let's perform the polynomial long division:

 x - 4
 x - 2 | x^2 - 6x + 10
 - (x^2 - 2x)
 ---------
 -4x + 10
 - (-4x + 8)
 ---------
 2

The result of the division is a quotient of x - 4 and a remainder of 2. This means we can express the original fraction as:

(x^2 - 6x + 10) / (x - 2) = (x - 4) + 2 / (x - 2)

Comparing this result with the given equation:

(x^2 - 6x + 10) / (x - 2) = A + 2 / (x - 2)

We can clearly see that A corresponds to the quotient we obtained from the long division, which is x - 4. This confirms our previous result and aligns with option (D).

This method of polynomial long division offers a different perspective on solving the problem. It highlights the relationship between polynomial division and the simplification of rational expressions. By understanding polynomial long division, students can tackle similar problems with confidence and gain a deeper appreciation for algebraic structures.

To further solidify our understanding, let's provide a more detailed explanation of the solution process. We started with the equation:

(x^2 - 6x + 10) / (x - 2) = A + 2 / (x - 2)

The goal is to find an expression for A. To achieve this, we need to isolate A on one side of the equation. The first step involves subtracting the term 2 / (x - 2) from both sides. This gives us:

(x^2 - 6x + 10) / (x - 2) - 2 / (x - 2) = A

Now, we have two fractions with the same denominator on the left side. This allows us to combine them into a single fraction. We do this by subtracting the numerators:

[(x^2 - 6x + 10) - 2] / (x - 2) = A

Simplifying the numerator involves combining like terms. In this case, we subtract 2 from 10:

(x^2 - 6x + 8) / (x - 2) = A

The next crucial step is to factor the quadratic expression in the numerator, x² - 6x + 8. Factoring a quadratic involves finding two binomials that, when multiplied, give us the original quadratic. We look for two numbers that multiply to 8 (the constant term) and add up to -6 (the coefficient of the x term). These numbers are -4 and -2. Therefore, we can factor the quadratic as:

x² - 6x + 8 = (x - 4)(x - 2)

Substituting this factored form back into our equation, we get:

[(x - 4)(x - 2)] / (x - 2) = A

Now, we have a common factor of (x - 2) in both the numerator and the denominator. We can cancel this factor, but it's important to remember that this cancellation is valid only when x ≠ 2, as the original expression is undefined at x = 2. After canceling the common factor, we are left with:

x - 4 = A

This is our final expression for A. It shows that A is equal to x - 4. This result corresponds to option (D) in the multiple-choice answers.

When tackling problems like this, it's easy to make mistakes if you're not careful. Here are some common pitfalls to watch out for:

  1. Incorrectly Combining Fractions: A common error is to incorrectly combine fractions. Remember, you can only add or subtract fractions directly if they have the same denominator. Ensure you find a common denominator before performing any operations on the numerators.
  2. Errors in Factoring: Factoring quadratics can be tricky. Double-check your factors by expanding them to ensure they match the original quadratic expression. A sign error or incorrect combination of factors can lead to a wrong answer.
  3. Forgetting to Consider Restrictions: The problem states that x ≠ 2. This restriction is crucial because it prevents division by zero. When simplifying expressions, be mindful of any restrictions on the variables and ensure your solution remains valid.
  4. Dividing Before Combining: Attempting to divide individual terms before combining the fractions can complicate the problem unnecessarily. Always simplify the expression by combining fractions first before attempting any division.
  5. Misinterpreting the Question: Make sure you understand what the question is asking. In this case, we needed to find an expression equivalent to A, not to solve for x. Misinterpreting the question can lead to irrelevant calculations and a wrong answer.

By being aware of these common mistakes, you can approach similar problems with greater confidence and accuracy.

While this problem might seem purely theoretical, the algebraic techniques it employs have numerous real-world applications. Understanding how to manipulate and simplify algebraic expressions is crucial in various fields, including:

  1. Engineering: Engineers use algebraic equations to model and analyze systems, design structures, and optimize processes. From calculating stress in a bridge to designing control systems for aircraft, algebra is a fundamental tool.
  2. Physics: Physics relies heavily on algebraic equations to describe the laws of nature. Equations of motion, energy conservation, and electromagnetism all involve algebraic manipulation and simplification.
  3. Computer Science: In computer programming, algebraic thinking is essential for developing algorithms and solving computational problems. Manipulating expressions, optimizing code, and understanding data structures all require algebraic skills.
  4. Economics: Economists use algebraic models to analyze markets, predict trends, and make policy recommendations. Supply and demand curves, economic growth models, and financial analysis all rely on algebraic techniques.
  5. Data Science: Data scientists use algebra to manipulate and analyze data, build machine learning models, and extract insights. Linear algebra, in particular, is a cornerstone of many data science algorithms.

The ability to simplify expressions, solve equations, and work with variables is a valuable skill that transcends the classroom. By mastering these techniques, you'll be well-equipped to tackle a wide range of challenges in various fields.

In conclusion, we have successfully solved the equation (x^2 - 6x + 10) / (x - 2) = A + 2 / (x - 2) and determined that A is equal to x - 4, corresponding to option (D). We explored two methods: algebraic manipulation and polynomial long division, both yielding the same result. Additionally, we discussed common mistakes to avoid and highlighted the real-world applications and significance of the algebraic techniques involved.

This problem serves as a valuable exercise in algebraic problem-solving, reinforcing the importance of careful manipulation, factoring, and simplification. By mastering these skills, students can confidently tackle a wide range of mathematical challenges and appreciate the power and versatility of algebra in various fields.

Algebraic equation, polynomial division, factoring, simplifying expressions, solving for A, mathematical problem, algebraic manipulation, real-world applications, common mistakes, detailed explanation.