Partial Fractions Revision Notes for AQA A-Level Mathematics

2.11.2 Squared Linear Denominators

When dealing with "squared linear denominators," you're working with algebraic fractions where the denominator is a squared linear expression, typically in the form $\ (axe + b)^2$ . These kinds of problems are common in calculus, algebra, and higher-level mathematics, and they require careful handling, especially when simplifying or solving equations.

Key Concepts

1. Simplifying Fractions with Squared Linear Denominators

When simplifying fractions that have squared linear denominators, you need to ensure that the numerator is in a suitable form that can potentially factor or simplify with the denominator.

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Example: Simplify $\frac{2x + 4}{(x + 2)^2}$ .

Solution:

Factor the numerator:

\frac{2(x + 2)}{(x + 2)^2}

Simplify by cancelling the common factor of $x + 2$ :

\frac{2}{x + 2}

So, the simplified fraction is $\frac{2}{x + 2}$ .

2. Adding and Subtracting Fractions with Squared Linear Denominators

To add or subtract fractions with different denominators, particularly when one or both are squared, you need to find a common denominator.

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Example: Add $\frac{1}{(x + 1)^2}$ and $\frac{2}{x + 1}$ .

Solution:

The common denominator is $(x + 1)^2$ .
Rewrite the second fraction with the common denominator:

\frac{2}{x + 1} = \frac{2(x + 1)}{(x + 1)^2}

Add the fractions:

\frac{1}{(x + 1)^2} + \frac{2(x + 1)}{(x + 1)^2} = \frac{1 + 2(x + 1)}{(x + 1)^2}

Simplify the numerator:

1 + 2(x + 1) = 1 + 2x + 2 = 2x + 3

So, the result is:

\frac{2x + 3}{(x + 1)^2}

3. Solving Equations Involving Squared Linear Denominators

When solving equations with squared linear denominators, you may need to multiply by the square of the linear expression to eliminate the fractions.

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Example: Solve $\frac{3}{(x - 2)^2} = \frac{4}{x - 2}$ .

Solution:

Multiply both sides by $(x - 2)^2$ to eliminate the denominators:

3 = 4(x - 2)

Expand and solve:

3 = 4x - 8

Rearrange to solve for $x$ :

4x = 11 \quad \Rightarrow \quad x = \frac{11}{4}

4. Partial Fractions with Squared Linear Denominators

Sometimes, you may encounter expressions where you need to decompose a fraction into partial fractions when the denominator is a squared linear expression.

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Example: Decompose $\frac{7x + 3}{(x + 2)^2}$ into partial fractions.

Solution:

Assume the form:

\frac{7x + 3}{(x + 2)^2} = \frac{A}{x + 2} + \frac{B}{(x + 2)^2}

Multiply through by $(x + 2)^2$ to get:

7x + 3 = A(x + 2) + B

Expand and equate coefficients:

7x + 3 = Axe + 2A + B

For $x$ -terms: $7 = A$
For constant terms: $3 = 2A + B$

Substitute $A = 7$ into the second equation:

3 = 2(7) + B \quad \Rightarrow \quad B = 3 - 14 = -11

So, the decomposition is:

\frac{7x + 3}{(x + 2)^2} = \frac{7}{x + 2} - \frac{11}{(x + 2)^2}

Practice Problem:

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Solve the equation $\frac{4}{(x - 3)^2} = \frac{1}{x - 3} + 3$ . Hint: Multiply through by $(x - 3)^2$ , then solve the resulting quadratic equation.

Solution:

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Step 1: Multiply through by $(x - 3)^2$ to eliminate the denominators:

4 = (x - 3) + 3(x - 3)^2

Step 2: Expand and simplify:

4 = x - 3 + 3(x^2 - 6x + 9)

4 = x - 3 + 3x^2 - 18x + 27

0 = 3x^2 - 17x + 20

Step 3: Solve the quadratic equation $3x^2 - 17x + 20 = 0$ using the quadratic formula:

x = \frac{-(-17) \pm \sqrt{(-17)^2 - 4(3)(20)}}{2(3)} = \frac{17 \pm \sqrt{49}}{6}

x = \frac{17 \pm 7}{6}

This gives the solutions:

x = 4 \quad \text{or} \quad x = \frac{5}{3}

These are the values of $x$ that satisfy the equation.

Partial Fractions (AQA A-Level Mathematics): Revision Notes

2.11.2 Squared Linear Denominators

Key Concepts

1. Simplifying Fractions with Squared Linear Denominators

2. Adding and Subtracting Fractions with Squared Linear Denominators

3. Solving Equations Involving Squared Linear Denominators

4. Partial Fractions with Squared Linear Denominators

Practice Problem:

Solution:

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