Problem 38 - Algebra A Combined Function

Question

Solve each equation. $$ \frac{1}{x+2}=\frac{4}{x^{2}-4}-\frac{1}{x-2} $$

Step-by-Step Solution

Answer

There is no valid solution for the given equation.

Step 1: Identify and Simplify Terms

The equation given is \( \frac{1}{x+2} = \frac{4}{x^2-4} - \frac{1}{x-2} \). The term \( x^2 - 4 \) can be factored as \( (x-2)(x+2) \). By factoring, we can identify common denominators for easier manipulation.

Step 2: Rewrite the Equation with Common Denominators

Rewrite each fraction with the common denominator \((x-2)(x+2)\):- \( \frac{1}{x+2} \) becomes \( \frac{x-2}{(x+2)(x-2)} \).- \( \frac{4}{x^2-4} \) is already over \((x+2)(x-2)\).- \( \frac{1}{x-2} \) becomes \( \frac{x+2}{(x-2)(x+2)} \).With these common denominators, the equation is \( \frac{x-2}{(x+2)(x-2)} = \frac{4}{x^2-4} - \frac{x+2}{(x-2)(x+2)} \).

Step 3: Combine Right-hand Side Fractions

Combine the fractions on the right side:\[ \frac{4}{(x-2)(x+2)} - \frac{x+2}{(x-2)(x+2)} = \frac{4 - (x+2)}{(x-2)(x+2)} = \frac{4-x-2}{(x-2)(x+2)} = \frac{2-x}{(x-2)(x+2)} \].

Step 4: Set Fractions Equal and Solve Numerators

Set the numerators equal since the denominators are the same:\( x-2 = 2-x \).Solve for \( x \) by adding \( x \) to both sides:\[ x + x - 2 = 2 \]\[ 2x - 2 = 2 \]Add 2 to both sides:\[ 2x = 4 \]Finally, divide by 2:\[ x = 2 \].

Step 5: Verify and Identify Constraints

Verify the solution is valid by checking it against the original denominators. The original denominators \( x+2 \) and \( x-2 \) must not be zero. However, setting \( x = 2 \) makes \( x-2 = 0 \), which is invalid. This means \( x=2 \) is not a part of the solution set.

Step 6: Conclude with No Valid Solution

Since the potential solution \( x = 2 \) leads to zero in one of the original denominators, we conclude that there is no value of \( x \) that satisfies the original equation.

Key Concepts

Factoring

Rational equations like these often include polynomials in the denominators, which can be tricky. Factoring is our secret weapon to simplify these equations. Take a look at the quadratic expression in the denominator: \(x^2 - 4\). This is a special case known as a difference of squares. Here’s how it works:

- The difference of squares formula is \(a^2 - b^2 = (a-b)(a+b)\).
- Recognize \(x^2 - 4\) as \(x^2 - 2^2\).
- You can factor it into \((x-2)(x+2)\).

This step is crucial because it shows us how \(x^2 - 4\) is related to its roots. Factoring simplifies the equation and enables us to find common denominators in the next steps.

Common Denominators

Common denominators are essential when working with rational equations. Let's see why. By rewriting each fraction to share a common denominator, we can combine the terms more easily.

- The common denominator for our problem is \((x-2)(x+2)\).
- For \(\frac{1}{x+2}\), we rewrite as \(\frac{x-2}{(x+2)(x-2)}\).
- For \(\frac{1}{x-2}\), rewrite as \(\frac{x+2}{(x-2)(x+2)}\).
- The expression \(\frac{4}{x^2-4}\) is already in the correct form.

This step is about aligning all parts of the equation. Once we have a common denominator across the fractions, we can efficiently combine and simplify them to solve for the variable. This is an indispensable skill in rational equations.

Solution Verification

After finding a solution, the final yet crucial step is to verify it. This ensures that your solution not only satisfies the equation but also doesn't invalidate any mathematical rules. Here's what you need to check:

- Ensure the solution does not make any denominator zero.
- In our case, substituting \(x = 2\) leads to \(x-2 = 0\), rendering the denominator invalid.
- Because floating-point operations can't break down in mathematics, you must exclude \(x = 2\) as a solution.

Verification is crucial because even mathematically valid steps can lead to answers that upset logical conditions, such as zero denominators. This session highlights the importance of not just finding solutions, but ensuring they're mathematically sound.