Problem 20 - Algebra A Combined Function

Question

Solve each equation and check each proposed solution. See Examples 4 through 6. $$ \frac{1}{x+2}+\frac{4}{x^{2}-4}=1 $$

Step-by-Step Solution

Answer

The solution is \(x = 3\).

Step 1: Recognize Factoring Opportunity

Identify that the denominator of the second fraction, \(x^2-4\), can be factored as \((x+2)(x-2)\). This factorization will help in finding a common denominator.

Step 2: Find Common Denominator

The least common denominator (LCD) for the fractions \(\frac{1}{x+2}\) and \(\frac{4}{x^2-4}\) is \((x+2)(x-2)\). Rewrite the equation with this common denominator: \(\frac{x-2 + 4}{(x+2)(x-2)} = 1\).

Step 3: Simplify the Numerator

Combine the terms in the numerator: \(x - 2 + 4 = x + 2\). Now the equation is \(\frac{x+2}{(x+2)(x-2)} = 1\).

Step 4: Cross-Multiply

To eliminate the fractions, cross-multiply to get \(x+2 = (x+2)(x-2)\).

Step 5: Simplify the Equation

Expand the right side of the equation: \(x+2 = x^2 - 4\).

Step 6: Move All Terms to One Side

Rearrange the equation into a standard quadratic form: \(x^2 - x - 6 = 0\).

Step 7: Solve the Quadratic Equation

Factor the quadratic equation: \((x-3)(x+2) = 0\). This gives potential solutions \(x = 3\) and \(x = -2\).

Step 8: Check Proposed Solutions

Substitute each proposed solution back into the original equation. \(x = -2\) makes the denominators undefined (division by zero). For \(x = 3\), the equation holds true as \(\frac{1}{5} + \frac{4}{5} = 1\).

Step 9: Conclude the Valid Solution

Since \(x = -2\) is invalid due to division by zero, the only valid solution is \(x = 3\).

Key Concepts

Factoring Polynomials

Factoring polynomials is a vital skill in algebra that helps solve complex equations more easily. When dealing with a polynomial expression, the goal is to rewrite it as a product of its simpler factors. For instance, consider the expression \(x^2 - 4\). This polynomial can be factored into \((x+2)(x-2)\).

This type of factorization is essential to identify when solving equations with polynomial denominators, as it allows us to simplify and combine terms. Here are some reasons why factoring polynomials is an important tool:

• Simplification: Factoring breaks down complicated expressions into simpler ones, making the equation easier to solve.
• Finding Zeros: Factoring helps find the values of \(x\) that make the equation equal to zero, known as the solutions or roots.
• Identifying Domain Restrictions: In equations with polynomial fractions, factoring helps identify values that might result in undefined expressions, like division by zero.

Factoring is not only a tool for solving equations but also a technique widely used across different areas of mathematics to simplify expressions.

Common Denominator

Finding a common denominator is a fundamental technique when working with fractions in algebra. A common denominator is a shared multiple of all the denominators in a set of fractions. In the equation \(\frac{1}{x+2} + \frac{4}{x^2-4} = 1\), recognizing that \(x^2 - 4\) can be factored into \((x+2)(x-2)\) allows us to establish a common denominator for these fractions.

Understanding how to find and use a common denominator is crucial because:

• Uniformity: It allows different fractions to be combined and manipulated as parts of a single equation.
• Simplification: Simplified expressions make it easier to apply operations like addition and subtraction across fractions.
• Elimination of fractions: Helps convert complex fractional equations into solvable integer expressions, simplifying the overall process.

After finding the least common denominator, you can rewrite each term with this new denominator, making it possible to combine them into a single equation.

Cross-Multiplication

Once you have rewritten your equation in terms of a single common denominator, you can sometimes simplify the process further by using cross-multiplication. Cross-multiplication is especially useful when dealing with fractions equal to each other or to an integer.

In our example, once the equation \(\frac{x+2}{(x+2)(x-2)} = 1\) is established, you can cross-multiply to eliminate the fractions on both sides. This means both sides of the equation are multiplied by the denominator to clear the fractions:

• This results in a simpler equation \(x+2 = (x+2)(x-2)\).
• Setting up the stage for solving the equation just like you would with any polynomial.

Cross-multiplying turns a potentially complex fractional equation into a manageable polynomial equation that can be worked with more easily.

Checking Solutions

Once the potential solutions of a quadratic equation are determined, the final step is to check these solutions by substituting them back into the original equation. Checking your solutions is crucial because:

• Verification: It ensures that the proposed solutions satisfy the original equation.
• Identifying Restrictions: Some solutions might introduce division by zero or other undefined expressions, such as \(x = -2\) in this case, making it an invalid solution.
• Accuracy: Verifying solutions helps avoid mistakes that might occur during the solving process.

In our exercise, substituting \(x = 3\) back into the equation confirms it as a correct solution, whereas \(x = -2\) is invalid as it leads to division by zero.