Problem 47
Question
write the partial fraction decomposition of each rational expression. $$ \frac{1}{x^{2}-c^{2}} \quad(c \neq 0) $$
Step-by-Step Solution
Verified Answer
The partial fraction decomposition of \( \frac{1}{x^{2}-c^{2}} \) is \( \frac{1}{2(x-c)} - \frac{1}{2(x+c)} \).
1Step 1: Identifying Difference of Squares
Firstly, identify that the denominator \( x^{2}-c^{2} \) is in the form of a difference of squares. This means it can be rewritten by using the identity \( a^{2} - b^{2} = (a-b)(a+b) \).
2Step 2: Breaking Down the Denominator
Applying the above identity to the denomination our expression, it becomes: \( \frac{1}{x^{2}-c^{2}} = \frac{1}{(x-c)(x+c)} \).
3Step 3: Writing as a Sum of Simpler Fractions
Now, the rule of partial fractions is that \( \frac{1}{(x-c)(x+c)} = \frac{A}{x-c} + \frac{B}{x+c} \), where A and B are constants. Multiplying all terms by the common denominator to clear the fractions, we get 1 = A(x+c) + B(x-c).
4Step 4: Finding the Constants A and B
To find A and B, two equations can be set up by choosing suitable values for x. Choosing \( x = c \) and \( x = -c \), we will get A = 1/2 and B = -1/2.
5Step 5: Presenting the final Answer
Substituting A and B back into the equation we get the final answer: \( \frac{1}{x^{2}-c^{2}} = \frac{1}{2(x-c)} - \frac{1}{2(x+c)} \).
Key Concepts
Difference of SquaresRational ExpressionsConstantsPartial Fractions
Difference of Squares
The term difference of squares refers to a specific kind of mathematical expression that can be rewritten using the formula \( a^2 - b^2 = (a-b)(a+b) \). Recognizing this pattern is crucial for simplifying expressions and solving equations efficiently.
This identity shows that a difference of squares can be factored into two binomials. Here:
This enables the expression \( x^2 - c^2 \) to be rewritten as \( (x-c)(x+c) \).
Spotting the difference of squares pattern is not always intuitive, but practicing it will help make calculations quicker and more straightforward.
This identity shows that a difference of squares can be factored into two binomials. Here:
- \( a \) represents \( x \)
- \( b \) stands for \( c \)
This enables the expression \( x^2 - c^2 \) to be rewritten as \( (x-c)(x+c) \).
Spotting the difference of squares pattern is not always intuitive, but practicing it will help make calculations quicker and more straightforward.
Rational Expressions
A rational expression is simply a fraction where the numerator and denominator are polynomials. Like regular fractions, they can be simplified, factored, and manipulated to make solving easier.
In this exercise, our rational expression is \( \frac{1}{x^2-c^2} \). To handle it effectively, identify any special forms like the difference of squares to simplify the expression. Once simplified, rational expressions can break down into simpler partial fractions which are easier to integrate or differentiate.
Understanding how to manipulate rational expressions by identifying common patterns, like factorable differences and similarities, aids in extracting meaningful solutions more conveniently.
In this exercise, our rational expression is \( \frac{1}{x^2-c^2} \). To handle it effectively, identify any special forms like the difference of squares to simplify the expression. Once simplified, rational expressions can break down into simpler partial fractions which are easier to integrate or differentiate.
Understanding how to manipulate rational expressions by identifying common patterns, like factorable differences and similarities, aids in extracting meaningful solutions more conveniently.
Constants
Constants in mathematical expressions are fixed values helping to solve unknowns in equations. In the realm of partial fraction decomposition, they play a vital role.
When decomposing \( \frac{1}{(x-c)(x+c)} \) into \( \frac{A}{x-c} + \frac{B}{x+c} \), \( A \) and \( B \) are the constants needing determination.
These constants are found by substituting strategic values for \( x \) that simplify the equation. As illustrated, choosing \( x = c \) solved for \( A \), and \( x = -c \) solved for \( B \). This method eliminates one constant at a time, making it easier to solve sequentially.
When decomposing \( \frac{1}{(x-c)(x+c)} \) into \( \frac{A}{x-c} + \frac{B}{x+c} \), \( A \) and \( B \) are the constants needing determination.
These constants are found by substituting strategic values for \( x \) that simplify the equation. As illustrated, choosing \( x = c \) solved for \( A \), and \( x = -c \) solved for \( B \). This method eliminates one constant at a time, making it easier to solve sequentially.
Partial Fractions
Partial fractions are a way of expressing a complicated rational expression as a sum of simpler fractions. This technique is particularly handy in calculus for integrating rational expressions.
In this exercise, \( \frac{1}{x^2-c^2} \) was rewritten as \( \frac{1}{(x-c)(x+c)} \), which can be further broken down into two simpler fractions: \( \frac{A}{x-c} + \frac{B}{x+c} \).
To determine \( A \) and \( B \), set up an equation such as \( 1 = A(x+c) + B(x-c) \) then solve for the constants as previously noted. Successfully doing so gives clarity and simplification to problems involving rational expressions.
This approach not only simplifies integration tasks but also makes complex algebraic manipulations more accessible.
In this exercise, \( \frac{1}{x^2-c^2} \) was rewritten as \( \frac{1}{(x-c)(x+c)} \), which can be further broken down into two simpler fractions: \( \frac{A}{x-c} + \frac{B}{x+c} \).
To determine \( A \) and \( B \), set up an equation such as \( 1 = A(x+c) + B(x-c) \) then solve for the constants as previously noted. Successfully doing so gives clarity and simplification to problems involving rational expressions.
This approach not only simplifies integration tasks but also makes complex algebraic manipulations more accessible.
Other exercises in this chapter
Problem 46
Graph the solution set of each system of inequalities or indicate that the system has no solution. $$ \left\\{\begin{array}{l} y \geq x^{2}-4 \\ x-y \geq 2 \end
View solution Problem 46
In Exercises \(43-46,\) let \(x\) represent one number and let \(y\) represent the other number. Use the given conditions to write a system of equations. Solve
View solution Problem 47
Graph the solution set of each system of inequalities or indicate that the system has no solution. $$ \left\\{\begin{array}{l} x^{2}+y^{2} \leq 16 \\ x+y>2 \end
View solution Problem 47
In Exercises \(47-48,\) solve each system by the method of your choice. $$ \left\\{\begin{array}{l} \frac{x+2}{2}-\frac{y+4}{3}=3 \\ \frac{x+y}{5}=\frac{x-y}{2}
View solution