Extraneous Solution Of 3/(a+2) + 2/a = (4a-4)/(a^2-4)

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Introduction

In the realm of algebra, solving equations is a fundamental skill. However, when dealing with rational equations, a unique challenge arises: the presence of extraneous solutions. These are solutions that emerge during the solving process but do not satisfy the original equation. Identifying and excluding these extraneous solutions is crucial for obtaining accurate results. In this comprehensive guide, we will delve into the intricacies of solving the rational equation $\frac{3}{a+2}+\frac{2}{a}=\frac{4 a-4}{a^2-4}$, pinpoint the extraneous solution, and provide a thorough explanation of the underlying concepts.

Understanding Rational Equations

Rational equations are equations that contain one or more rational expressions, which are fractions with polynomials in the numerator and/or denominator. Solving these equations involves manipulating the fractions to eliminate the denominators and obtain a simpler equation that can be solved using standard algebraic techniques. However, a critical step in this process is identifying any values that would make the denominators zero, as these values are excluded from the domain of the equation and may lead to extraneous solutions.

The given equation, $\frac{3}{a+2}+\frac{2}{a}=\frac{4 a-4}{a^2-4}$, is a prime example of a rational equation. To solve it effectively, we must first understand the nuances of working with rational expressions and the potential pitfalls of extraneous solutions.

Step-by-Step Solution

To solve the equation $\frac{3}{a+2}+\frac{2}{a}=\frac{4 a-4}{a^2-4}$, we will follow a systematic approach:

1. Identify Restricted Values

The first step is to identify any values of 'a' that would make the denominators of the fractions equal to zero. These values are restricted from the domain of the equation because division by zero is undefined. In this case, the denominators are $a+2$, $a$, and $a^2-4$. Setting each of these equal to zero, we get:

• $a+2=0 \implies a=-2$
• $a=0$
• $a^2-4=0 \implies (a+2)(a-2)=0 \implies a=-2$ or $a=2$

Therefore, the restricted values are $a=-2$, $a=0$, and $a=2$. These values cannot be solutions to the equation.

2. Find the Least Common Denominator (LCD)

To eliminate the fractions, we need to find the LCD of the denominators. The denominators are $a+2$, $a$, and $a^2-4$, which can be factored as $(a+2)(a-2)$. The LCD is the product of the unique factors, each raised to the highest power that appears in any of the denominators. In this case, the LCD is $a(a+2)(a-2)$.

3. Multiply Both Sides by the LCD

Multiply both sides of the equation by the LCD to eliminate the fractions:

$a(a+2)(a-2) \left( \frac{3}{a+2}+\frac{2}{a} \right) = a(a+2)(a-2) \left( \frac{4 a-4}{a^2-4} \right)$

Distribute the LCD to each term:

$3a(a-2) + 2(a+2)(a-2) = (4a-4)a$

4. Simplify and Solve the Equation

Expand and simplify the equation:

$3a^2 - 6a + 2(a^2 - 4) = 4a^2 - 4a$

$3a^2 - 6a + 2a^2 - 8 = 4a^2 - 4a$

Combine like terms:

$5a^2 - 6a - 8 = 4a^2 - 4a$

Subtract $4a^2 - 4a$ from both sides:

$a^2 - 2a - 8 = 0$

Factor the quadratic equation:

$(a-4)(a+2) = 0$

Set each factor equal to zero and solve for 'a':

$a-4=0 \implies a=4$

$a+2=0 \implies a=-2$

5. Check for Extraneous Solutions

Now, we need to check if the solutions we found are extraneous by plugging them back into the original equation. Remember that we identified $a=-2$, $a=0$, and $a=2$ as restricted values.

For $a=4$:

$\frac{3}{4+2}+\frac{2}{4}=\frac{4(4)-4}{4^2-4}$

$\frac{3}{6}+\frac{1}{2}=\frac{16-4}{16-4}$

$\frac{1}{2}+\frac{1}{2}=\frac{12}{12}$

$1=1$

So, $a=4$ is a valid solution.

For $a=-2$:

$\frac{3}{-2+2}+\frac{2}{-2}=\frac{4(-2)-4}{(-2)^2-4}$

$\frac{3}{0}+\frac{2}{-2}=\frac{-8-4}{4-4}$

Since we have division by zero, $a=-2$ is an extraneous solution.

Identifying Extraneous Solutions

The solution $a=-2$ is extraneous because it makes the denominators of the original equation equal to zero, which is undefined. This highlights a critical aspect of solving rational equations: always check your solutions against the restricted values to identify and exclude any extraneous solutions.

Why Extraneous Solutions Occur

Extraneous solutions arise when we perform operations that are not reversible in the context of the original equation. In the case of rational equations, multiplying both sides by the LCD can introduce solutions that satisfy the transformed equation but not the original equation. This is because the LCD might be zero for certain values of the variable, and multiplying by zero can make unequal quantities appear equal.

Conclusion

In the given equation $\frac{3}{a+2}+\frac{2}{a}=\frac{4 a-4}{a^2-4}$, the extraneous solution is a = -2. This solution was generated during the algebraic manipulation but does not satisfy the original equation because it results in division by zero. The valid solution is a = 4.

This exercise underscores the importance of identifying restricted values and checking for extraneous solutions when solving rational equations. By following a systematic approach and being mindful of these potential pitfalls, you can confidently navigate the complexities of rational equations and arrive at accurate solutions.

Understanding extraneous solutions is paramount in solving rational equations. Always remember to check your solutions against the original equation to ensure their validity. This step is crucial for avoiding errors and obtaining correct results in algebraic problem-solving. In summary, the extraneous solution in the given equation is $a = -2$, a value that makes the denominator zero and renders the equation undefined.

Final Answer

The extraneous solution to the equation $\frac{3}{a+2}+\frac{2}{a}=\frac{4 a-4}{a^2-4}$ is A. a = -2.