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

Jul 7, 2025 by Admin 76 views

Which Solution to the Equation 3/(a+2) + 2/a = (4a-4)/(a^2-4) is Extraneous? A Comprehensive Guide

#h1 Understanding Extraneous Solutions in Rational Equations

When dealing with rational equations, it's crucial to understand the concept of extraneous solutions. These are solutions that arise during the solving process but do not satisfy the original equation. They often occur when we perform operations that can introduce values that make the denominators zero, which is undefined in mathematics. In this comprehensive guide, we will delve into the solution of the equation $\frac{3}{a+2} + \frac{2}{a} = \frac{4a-4}{a^2-4}$ , identify the extraneous solution, and provide a step-by-step explanation to help you grasp the underlying principles. The question at hand asks us to identify the extraneous solution of the equation, meaning we need to find the value of 'a' that satisfies the transformed equation but not the original one. This typically happens when we eliminate denominators or perform operations that alter the domain of the equation. Therefore, a meticulous approach is required to solve the equation and subsequently check for extraneous solutions. Our goal is not just to find the correct answer but also to understand why certain solutions are valid while others are not. This understanding is fundamental in solving rational equations and avoiding common pitfalls. By the end of this guide, you will be equipped with the knowledge and skills to confidently tackle similar problems. We will break down the problem into manageable steps, explaining each operation and its implications. Remember, the key to solving rational equations lies in identifying the restrictions on the variable and verifying solutions against the original equation.

#h2 Solving the Rational Equation Step-by-Step

Let's start by solving the equation: $\frac{3}{a+2} + \frac{2}{a} = \frac{4a-4}{a^2-4}$ .

1. Factoring the Denominator

The first step in solving any rational equation is to factor the denominators. This helps us identify common factors and determine the least common denominator (LCD). In this case, we can factor $a^2 - 4$ as a difference of squares:

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

Now our equation looks like this:

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

2. Identifying Restrictions

Before proceeding further, it's imperative to identify the restrictions on the variable 'a'. These are values of 'a' that would make any of the denominators zero, rendering the expression undefined. From the denominators $a+2$ , $a$ , and $(a+2)(a-2)$ , we can see that:

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

So, our restrictions are $a \neq -2$ , $a \neq 0$ , and $a \neq 2$ . These values cannot be solutions to the original equation.

3. Finding the Least Common Denominator (LCD)

The LCD is the smallest expression that is divisible by all the denominators. In this case, the LCD is $a(a+2)(a-2)$ .

4. Multiplying by the LCD

To eliminate the fractions, we multiply both sides of the equation by the LCD:

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

Distributing the LCD on the left side, we get:

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

5. Expanding and Simplifying

Now, let's expand and simplify the equation:

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

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

Combining like terms, we have:

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

6. Rearranging the Equation

Move all terms to one side to set the equation to zero:

$5a^2 - 6a - 8 - 4a^2 + 4a = 0$

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

7. Factoring the Quadratic

Factor the quadratic equation:

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

8. Solving for 'a'

Setting each factor equal to zero, we find the potential solutions:

$a - 4 = 0 \implies a = 4$
$a + 2 = 0 \implies a = -2$

#h3 Identifying the Extraneous Solution

We have found two potential solutions: $a = 4$ and $a = -2$ . However, we must check these solutions against the restrictions we identified earlier. Remember, our restrictions were $a \neq -2$ , $a \neq 0$ , and $a \neq 2$ .

Checking a = 4

$a = 4$ does not violate any of our restrictions, so it is a valid solution. We can verify this by plugging $a = 4$ back into the original equation:

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

$\frac{3}{6} + \frac{1}{2} = \frac{12}{12}$

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

$1 = 1$

This confirms that $a = 4$ is a valid solution.

Checking a = -2

$a = -2$ violates the restriction $a \neq -2$ . Therefore, $a = -2$ is an extraneous solution. Plugging $a = -2$ into the original equation would result in division by zero, which is undefined.

#h4 Conclusion: The Extraneous Solution

In conclusion, the extraneous solution to the equation $\frac{3}{a+2} + \frac{2}{a} = \frac{4a-4}{a^2-4}$ is $a = -2$ . This is because, while it satisfies the simplified quadratic equation, it makes the denominator of the original equation equal to zero, which is not permissible. Therefore, the correct answer is:

A. $a = -2$

#h5 Key Takeaways for Solving Rational Equations

Factor the denominators: This simplifies the equation and helps identify the LCD.
Identify restrictions: Determine the values of the variable that make the denominators zero. These values cannot be valid solutions.
Find the LCD: This allows you to eliminate the fractions by multiplying both sides of the equation.
Multiply by the LCD: This step clears the fractions, making the equation easier to solve.
Expand and simplify: Combine like terms to obtain a simpler equation.
Solve the resulting equation: This may involve factoring, using the quadratic formula, or other algebraic techniques.
Check for extraneous solutions: Always compare your solutions to the restrictions. Any solution that violates a restriction is extraneous.

By following these steps, you can confidently solve rational equations and avoid the common pitfall of extraneous solutions. Remember, the key is to be meticulous and check your work at each step. Understanding why a solution is extraneous is as important as finding the solution itself.

#h6 Practice Problems

To solidify your understanding, try solving these practice problems:

$\frac{1}{x-1} + \frac{1}{x+1} = \frac{2}{x^2-1}$
$\frac{2}{x+3} - \frac{1}{x} = \frac{-3}{x(x+3)}$
$\frac{x}{x-2} + \frac{2}{x+2} = \frac{8}{x^2-4}$

Work through these problems, paying close attention to identifying restrictions and checking for extraneous solutions. The more you practice, the more comfortable you will become with solving rational equations.

By mastering the techniques discussed in this guide, you will be well-prepared to tackle a wide range of rational equation problems. Remember to always check your solutions and understand the concept of extraneous solutions to ensure accurate results. Rational equations may seem daunting at first, but with a systematic approach and careful attention to detail, they can be solved with confidence.