Problem 13 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-14: Solve the equation by factoring. $$ \frac{5 x}{x-3}+\frac{4}{x+3}=\frac{90}{x^{2}-9} $$

Step-by-Step Solution

Answer

Solutions are \(x = \frac{6}{5}\) and \(x = -17\).

Step 1: Identify the Problem

We need to solve the equation by factoring: \(\frac{5x}{x-3} + \frac{4}{x+3} = \frac{90}{x^2-9}\).

Step 2: Factor the Denominator

Recognize that \(x^2 - 9\) is a difference of squares and factor it as \((x-3)(x+3)\). So, \(\frac{90}{x^2-9} = \frac{90}{(x-3)(x+3)}\).

Step 3: Find a Common Denominator

The common denominator for the fractions \(\frac{5x}{x-3}\), \(\frac{4}{x+3}\), and \(\frac{90}{(x-3)(x+3)}\) is \((x-3)(x+3)\).

Step 4: Adjust Fractions to Common Denominator

Rewrite the equation with the common denominator:\[ \frac{5x(x+3)}{(x-3)(x+3)} + \frac{4(x-3)}{(x-3)(x+3)} = \frac{90}{(x-3)(x+3)} \]

Step 5: Combine and Simplify the Numerators

Combine the numerators into a single expression:\[ 5x(x+3) + 4(x-3) = 90 \] Expand and simplify:\[ 5x^2 + 15x + 4x - 12 = 90 \] So, \[ 5x^2 + 19x - 102 = 0 \].

Step 6: Factor the Quadratic Equation

To factor \(5x^2 + 19x - 102 = 0\), split the middle term:\[ (5x - 6)(x + 17) = 0 \].

Step 7: Solve for x

Set each factor equal to zero:1. \( 5x - 6 = 0 \) leads to \( x = \frac{6}{5} \).2. \( x + 17 = 0 \) leads to \( x = -17 \).

Step 8: Check for Extraneous Solutions

Substitute \( x = \frac{6}{5} \) and \( x = -17 \) back into the original denominators \(x-3\) and \(x+3\) to ensure they do not cause division by zero. Both solutions are valid.

Key Concepts

Difference of Squares

The difference of squares is a powerful tool in algebra, especially when it comes to factoring. It refers to the expression in the form of \( a^2 - b^2 \). This can be factored into \((a-b)(a+b)\).
In our equation \( x^2 - 9 \), we recognize this as a difference of squares because it can be written as \( x^2 - 3^2 \). Hence, it factors into \((x-3)(x+3)\).
Once identified, factoring the difference of squares helps simplify equations and solve them more efficiently. Always look for patterns like \( a^2 - b^2 \) in polynomials. They often hide in higher-level algebra problems and can be your key to simplifying complex expressions.

Common Denominator

When solving equations with fractions, finding a common denominator is essential. This allows you to combine fractions easily.
In the given exercise, fractions \( \frac{5x}{x-3} \), \( \frac{4}{x+3} \), and \( \frac{90}{x^2 - 9} \) need a common base. The expression \( x^2 - 9 \) can be factored as a difference of squares \((x-3)(x+3)\). Hence, this serves as the common denominator.
The key steps:

• Rewrite each fraction with the common denominator: \( (x-3)(x+3) \).
• Ensure all numerators are adjusted accordingly by multiplying with terms to reach the common denominator.
• This helps combine all fractions into a single equation, making it easier to solve.

Proper identification and manipulation of the common denominator are crucial for solving such equations efficiently.

Quadratic Equation

A quadratic equation is any equation in the form of \( ax^2 + bx + c = 0 \). Solving them often involves factoring, completing the square, or using the quadratic formula.
In the exercise, after adopting a common denominator, we combined the fractions into an expanded quadratic equation \( 5x^2 + 19x - 102 = 0 \).
Here's a quick guide to factoring quadratic equations:

• Look for two numbers that multiply to give \( ac \) and add to give \( b \).
• Split the middle term, using these figures.
• Factor by grouping, to express the quadratic equation as a product of two binomials.

In this exercise, factoring gave the binomials \((5x - 6)(x + 17)\), leading to the solutions \( x = \frac{6}{5} \) and \( x = -17 \).

Extraneous Solutions

Extraneous solutions are roots that emerge from the solving process but don't satisfy the original equation. They often arise when dealing with rational expressions.
After finding solutions to an equation, always substitute back to check validity, particularly when fractions are involved.
For instance, in this problem:

• We obtained solutions \( x = \frac{6}{5} \) and \( x = -17 \).
• Substituting back, both solutions do not make any denominator zero, thus they are both valid.

Beware of extraneous solutions as they can mislead you. Always re-check with the original equation to confirm that these solutions work, particularly when rational expressions are involved.