Question

Resolve \(\frac{x-b}{x^{2}+(a+b) x+a b}\) into partial fractions. (1) \(\frac{2 a}{(a+b)(x+a)}-\frac{2 b}{(a+b)(x+b)}\) (2) \(\frac{a+b}{(a-b)(x+a)}+\frac{2 b}{(a-b)(x+b)}\) (3) \(\frac{a+b}{(a-b)(x+a)}-\frac{2 b}{(a-b)(x+b)}\) (4) None of these

Short answer

Answer: None of these. The correct partial fraction decomposition is \(\frac{(a-b)}{a(x+a)}-\frac{(a+b)}{2a(x+b)}\).

Step-by-step solution

Write the given fraction as the sum of partial fractions

The general form for a quadratic decomposition will be: \(\frac{x-b}{x^{2}+(a+b) x+a b} = \frac{A}{x+a}+\frac{B}{x+b}\) Now our goal is to find values of A and B.

Clear the denominators

Multiply both sides by \((x+a)(x+b)\) to clear the denominators. We get: \((x-b) = A(x+b) + B(x+a)\)

Solve for A and B

At this stage, we have a linear system of equations to solve for A and B. We can use substitution or elimination. In this case, let's use substitution by plugging particular values for x: 1. Plug \(x=-a\) to solve for B: \((-a-b) = B(-2a)\) \(B = \frac{a+b}{-2a}\) 2. Plug \(x=-b\) to solve for A: \((a-b) = A(a)\) \(A = \frac{a-b}{a}\) Now we have found A and B: \(A = \frac{a-b}{a}\) \(B = \frac{a+b}{-2a}\)

Put the values of A and B back into the partial fractions

Now that we have the values of A and B, let's substitute them back into the partial fraction decomposition: \(\frac{x-b}{x^{2}+(a+b) x+a b} = \frac{a-b}{a(x+a)}+\frac{a+b}{-2a(x+b)}\) Now simplify the expression: \(\frac{x-b}{x^{2}+(a+b) x+a b} = \frac{(a-b)}{a(x+a)}-\frac{(a+b)}{2a(x+b)}\) For the given options, we can write our expression as: \(\frac{x-b}{x^{2}+(a+b) x+a b} = \frac{2(a-b)}{2a(x+a)}-\frac{2(a+b)}{2a(x+b)}\) Comparing with the given options, it can be seen that none of them match with our final answer, so the correct choice is: (4) None of these

Key concepts

Algebraic Fractions

Algebraic fractions are fractions where the numerator, the denominator, or both contain algebraic expressions. They are similar to numerical fractions, but instead of numbers, they have variables and constants. The process of simplifying algebraic fractions is essential, as it allows us to find simpler forms or break them into smaller parts, making them easier to work with.

In the given exercise, we work with an algebraic fraction with a quadratic denominator. For simplification, we use partial fraction decomposition, an effective method that expresses a complex fraction as a sum of simpler fractions, known as partial fractions. These partial fractions have denominators that are factors of the original denominator. Appropriately choosing and solving for constants in these partial fractions helps us to represent complex algebraic relationships in a more manageable way.

Linear System of Equations

A linear system of equations consists of multiple linear equations that can be solved to find the values of unknown variables. When dealing with partial fraction decomposition, we encounter linear systems in the process of solving for the coefficients of the simpler fractions.

After clearing denominators and equating coefficients for the original algebraic fraction and the proposed partial fractions, we end up with a linear system. This system is then solved using well-established methods like substitution, elimination, or matrix operations. In our exercise, substitution was chosen to find the values of A and B. This approach is straightforward and efficient for systems with a small number of equations and unknowns. It involves solving one equation for one variable and then using that result to solve for another variable, iteratively reducing the complexity until all values are found.

Quadratic Decomposition

Quadratic decomposition is a technique used to break down a fraction with a quadratic polynomial in the denominator into simpler fractions with linear denominators. The idea is to take a complex fraction and decompose it into a sum of fractions whose denominators are the factors of the original quadratic.

During the process, we assume a general form for the partial fractions, where each fraction has an unknown constant to be determined. The goal is to find these constants such that when the partial fractions are combined, they equate to the original algebraic fraction. In the problem presented, the quadratic denominator is factored into two linear factors, leading to a decomposition involving constants A and B. The process of solving for these constants is closely tied to solving a linear system of equations, ultimately providing the coefficients for the decomposed linear fractions.

The exercise improvement advice suggests ensuring a clear explanation of each step: breaking down the quadratic, assuming general forms for the constants, and methodically solving the linear system to find their values. A clear and patient approach to each of these steps is vital for a deep understanding of quadratic decomposition.