Extraneous Solution In The Equation 3/(a+2) + 2/a = (4a-4)/(a^2-4)

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In this comprehensive guide, we will delve into the intricacies of solving the equation 3/(a+2) + 2/a = (4a-4)/(a^2-4) and, more importantly, identifying any extraneous solutions that may arise. Extraneous solutions are values that satisfy the transformed equation but not the original equation due to operations performed during the solving process, such as squaring or multiplying by an expression containing the variable. Understanding how to pinpoint these extraneous solutions is crucial for ensuring the accuracy of our mathematical results. We will explore the step-by-step process of solving the equation, paying close attention to the domain restrictions imposed by the denominators and how these restrictions help us identify extraneous solutions. This detailed exploration will empower you to confidently tackle similar problems and deepen your understanding of algebraic equation solving.

Before we dive into the specific equation at hand, let's first clarify what extraneous solutions are and why they occur. In essence, an extraneous solution is a solution that emerges during the process of solving an equation but does not actually satisfy the original equation. This often happens when we perform operations that can introduce new solutions that weren't there initially. Common operations that can lead to extraneous solutions include squaring both sides of an equation, multiplying both sides by an expression containing a variable, or, as in our case, dealing with rational equations where variables appear in the denominators.

In the context of rational equations, extraneous solutions typically arise when a value of the variable makes one of the denominators in the original equation equal to zero. Since division by zero is undefined, these values are not permissible solutions, even if they satisfy the transformed equation after we've cleared the fractions. Therefore, it's imperative to identify these restricted values before we begin solving and to check our final solutions against these restrictions. Failing to do so can lead to incorrect answers and a misunderstanding of the solution set. The process of identifying and excluding extraneous solutions is a critical step in solving rational equations accurately.

To solve the equation 3/(a+2) + 2/a = (4a-4)/(a^2-4), we will follow a series of algebraic steps. Our primary goal is to isolate the variable a and determine the values that satisfy the equation. However, before we embark on this journey, it's absolutely crucial to identify any values of a that would make the denominators of our fractions equal to zero. These values will be our potential extraneous solutions, and we must keep them in mind as we proceed.

Step 1: Identify Restricted Values

We begin by examining the denominators of our fractions: (a+2), a, and (a^2-4). Setting each of these equal to zero, we find the values of a that would make them zero:

  • a + 2 = 0 => a = -2
  • a = 0
  • a^2 - 4 = 0 => (a+2)(a-2) = 0 => a = -2 or a = 2

Thus, we have three restricted values: a = -2, a = 0, and a = 2. These values cannot be solutions to our original equation because they would result in division by zero, which is undefined. Any solutions we find that match these values will be classified as extraneous.

Step 2: Find the Least Common Denominator (LCD)

To eliminate the fractions, we need to multiply both sides of the equation by the least common denominator (LCD) of the fractions. The denominators are (a+2), a, and (a^2-4), which can be factored as (a+2)(a-2). The LCD is therefore the product of the unique factors, each raised to the highest power it appears in any denominator. In this case, the LCD is a(a+2)(a-2).

Step 3: Multiply Both Sides by the LCD

Multiplying both sides of the equation by the LCD, a(a+2)(a-2), we get:

  • a(a+2)(a-2) * [3/(a+2) + 2/a] = a(a+2)(a-2) * [(4a-4)/(a^2-4)]

Now, we distribute the LCD to each term:

  • 3a(a-2) + 2(a+2)(a-2) = (4a-4)*a

Step 4: Simplify and Solve the Equation

Expanding and simplifying the equation, we have:

  • 3a^2 - 6a + 2(a^2 - 4) = 4a^2 - 4a
  • 3a^2 - 6a + 2a^2 - 8 = 4a^2 - 4a
  • 5a^2 - 6a - 8 = 4a^2 - 4a

Moving all terms to one side to set the equation to zero, we get:

  • a^2 - 2a - 8 = 0

Factoring the quadratic equation, we have:

  • (a - 4)(a + 2) = 0

This gives us two potential solutions:

  • a = 4 or a = -2

Step 5: Check for Extraneous Solutions

Now, we must check our potential solutions against the restricted values we identified earlier (a = -2, a = 0, and a = 2). We found potential solutions of a = 4 and a = -2.

  • a = 4: This value is not among our restricted values, so it is a valid solution.
  • a = -2: This value is a restricted value because it makes the denominators (a+2) and (a^2-4) equal to zero. Therefore, a = -2 is an extraneous solution.

After meticulously solving the equation 3/(a+2) + 2/a = (4a-4)/(a^2-4), we arrived at two potential solutions: a = 4 and a = -2. However, our analysis of the restricted values revealed that a = -2 is an extraneous solution. This is because substituting a = -2 into the original equation results in division by zero, rendering the equation undefined. In contrast, a = 4 does not violate any domain restrictions and, therefore, is a valid solution. This detailed process highlights the critical importance of checking for extraneous solutions when solving rational equations.

In conclusion, the extraneous solution to the equation 3/(a+2) + 2/a = (4a-4)/(a^2-4) is a = -2. This example underscores the importance of identifying restricted values and checking for extraneous solutions when working with rational equations. By carefully considering the domain restrictions and verifying our solutions, we can ensure the accuracy and validity of our mathematical results. This understanding is not only crucial for solving equations but also for developing a deeper appreciation for the nuances of algebraic problem-solving. Remember to always be vigilant in identifying potential extraneous solutions to ensure the correctness of your answers.