Problem 107 - Algebra and Trigonometry

Question

Solve equation by the method of your choice. $$ \frac{2 x}{x-3}+\frac{6}{x+3}=-\frac{28}{x^{2}-9} $$

Step-by-Step Solution

Answer

The solution to the equation is x = -5

Step 1: Identifying the Common Denominator

Find the common denominator. In this case, the common denominator for all the fractions is \(x^{2}-9\) since it can be factored to \((x-3)(x+3)\). So, multiply each term with a value that turns the denominator of each term to \(x^{2}-9\). This will lead to: \[2x(x+3) + 6(x-3) = -28\].

Step 2: Simplify the Equation

Using the distributive property, combine the terms in parentheses and simplify further by moving terms to one side to arrive at a quadratic equation: \[2x^{2} +6x +6x -18 + 28 = 0 \Rightarrow 2x^{2}+12x+10=0\]. Dividing throughout by 2 leads to \(x^{2} + 6x + 5 = 0\). This becomes a simple quadratic equation.

Step 3: Solve the Quadratic Equation

A quadratic equation of the form \(ax^2 + bx + c = 0\) can be solved using the quadratic formula \(x = [-b ± sqrt(b^2 - 4ac)] / 2a\). Applying this formula here, the roots are \[x = [-6 ± sqrt((6)^2 - 4*5)] / 2 = -1, -5\]. The solutions need to make the original denominators not equal zero. Checking the solutions, we observe that only x = -5 is valid since x = -1 causes a zero denominator in the original equation. Thus, x = -5 is the solution.

Key Concepts

Common Denominator

When solving equations that involve fractions, finding a common denominator is essential. A common denominator is a shared multiple of the denominators in different fractions, allowing you to combine them into a single fraction. In our original exercise, the denominators were \(x-3\), \(x+3\), and \(x^2-9\), which can be factored into \( (x-3)(x+3) \).
This means that \(x^2-9\) is the least common denominator that can encompass all fractions in the equation. By multiplying each term by this common denominator, we eliminate the fractional parts, simplifying the equation. This step is crucial to solving the equation effectively.

Quadratic Equation

A quadratic equation is a type of polynomial equation of degree two, typically expressed in the form \(ax^2 + bx + c = 0\). In our solution, after clearing the fractions, we arrived at the equation \(2x^2 + 12x + 10 = 0\).
Quadratic equations often have two solutions, which can be real or complex numbers. In this specific exercise, simplifying the coefficients further by dividing through by 2 gives us \(x^2 + 6x + 5 = 0\).
Recognizing this as a standard quadratic equation allows us to apply standard solving methods, such as factoring, completing the square, or using the quadratic formula.

Distributive Property

Understanding the distributive property is vital when working with polynomials and equations. The distributive property states that \(a(b + c) = ab + ac\), which means you distribute the multiplication over addition or subtraction within parentheses.
In our exercise, applying the distributive property allows us to expand terms like \(2x(x+3) + 6(x-3)\). By distributing, we arrive at \(2x^2 + 6x + 6x - 18\). Simplifying these terms by combining like terms and moving others helps transform the equation into a manageable quadratic form. This step is a foundational skill in algebra, providing a pathway to solve complex equations.

Quadratic Formula

The quadratic formula is a powerful tool used to solve quadratic equations of the form \(ax^2 + bx + c = 0\). The formula is given by \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]. This formula provides the roots of the quadratic equation as a solution.
In our example, after simplifying the equation to \(x^2 + 6x + 5 = 0\), we used the quadratic formula. Plugging in the coefficients \(a = 1\), \(b = 6\), and \(c = 5\), we found the solutions \(x = -1\) and \(x = -5\).
However, only \(x = -5\) was valid, as it does not cause a zero denominator in the original equation. Mastering the quadratic formula is essential for solving quadratic equations when factors are not easily recognizable.