Problem 27 - Beginning and Intermediate Algebra

Question

Solve. $$\frac{x}{x-4}-\frac{4}{x+4}=\frac{32}{x^{2}-16}$$

Step-by-Step Solution

Answer

The short answer to solving the given equation \(\frac{x}{x-4}-\frac{4}{x+4}=\frac{32}{x^{2}-16}\) is: after finding the common denominator, simplifying, and solving for x, we find \(x = ±4\). However, both of these solutions are extraneous, as they result in undefined fractions. Therefore, there are no valid solutions for this equation.

Step 1: Find the common denominator

To find the common denominator for \(\frac{x}{x-4}\), \(\frac{4}{x+4}\), and \(\frac{32}{x^2 - 16}\), we first factor the denominators: \((x-4)\), \((x+4)\), and \((x-4)(x+4)\). The common denominator for these fractions will be the least common multiple of the denominators, which is \((x-4)(x+4)\).

Step 2: Rewrite each fraction with the common denominator

Next, rewrite each fraction with the common denominator \((x-4)(x+4)\): \[\frac{x(x+4)}{(x-4)(x+4)} - \frac{4(x-4)}{(x+4)(x-4)} = \frac{32}{(x-4)(x+4)}\]

Step 3: Simplify and remove the denominators

Now, we can use the common denominator to simplify and remove the denominators from the equation: \[(x(x+4)) - 4(x-4) = 32\]

Step 4: Expand and solve for x

Expand the equation and group the terms: \[x^2 + 4x - 4x + 16 = 32\] Simplify further: \[x^2 = 16\] To solve for x, take the square root of both sides: \[x = ±4\]

Step 5: Check for extraneous solutions

Finally, substitute these values of x back into the original equation and check for extraneous solutions. If \(x = -4\) or \(x = 4\), one of the denominators becomes zero, which results in an undefined fraction. Therefore, both \(x = -4\) and \(x = 4\) are extraneous solutions, and there are no valid solutions for this equation.

Key Concepts

Common Denominator

When solving rational equations, finding a common denominator is an essential step. It allows us to combine fractions into a single equation without changing their values.

In our example, the fractions \(\frac{x}{x-4}\), \(\frac{4}{x+4}\), and \(\frac{32}{x^2 - 16}\) need a common denominator. We first factor the denominators to find this shared base.

Here are the steps:

• Factor each denominator: \((x-4)\), \((x+4)\), and \((x-4)(x+4)\).
• Identify the least common multiple (LCM): \((x-4)(x+4)\).
• Rewrite each fraction using this common denominator.

This process ensures all parts of the equation are expressed over a unified denominator, allowing us to effectively simplify and solve the equation.

Extraneous Solutions

Extraneous solutions are results of solving equations that do not satisfy the original equation. This usually happens when solutions make any denominator equal to zero, leading to an undefined expression.

In the solution for our equation, we derived \(x = ±4\) as potential solutions. However, substituting these back into the original equation shows a problem:

• \(x = 4\) leads to a zero denominator in \(\frac{x}{x-4}\) and \(x = -4\) leads to a zero denominator in \(\frac{4}{x+4}\).

Since these values make parts of the equation undefined, \(x = 4\) and \(x = -4\) are extraneous. It's essential to check potential solutions by plugging them back into the original equation to verify their validity.

Factoring Polynomials

Factoring polynomials is a critical step in simplifying rational equations. It involves breaking down a polynomial into a product of simpler expressions, which helps in finding common denominators.

In our exercise, the polynomial \((x^2 - 16)\) was factored to \((x-4)(x+4)\). This is a difference of squares, a common type of polynomial factorization:

• The difference of squares formula is \(a^2 - b^2 = (a-b)(a+b)\).
• Using this identity makes it easier to manage the polynomial, especially when finding least common multiples or simplifying expressions.

Mastering the skill of factoring polynomials can greatly ease the process of solving rational equations.

Quadratic Equations

Quadratic equations form a significant part of solving rational equations, especially when simplifying or solving the expressions leads to a quadratic form.

In our example, the simplification led us to \(x^2 = 16\). This is a basic quadratic equation, which we solve by taking the square root of both sides.

The steps involved include:

• Recognizing and rewriting the expression as a perfect square: \(x^2 = 16\).
• Applying the square root to both sides: \(x = ±4\).

While solving these, it's important to consider potential extraneous solutions, as demonstrated in our exercise, to ensure the solutions obtained are valid.