Algebra Revision Notes

Basic Steps for x² + bx + c

To complete the square for a quadratic expression of the form x² + bx + c:

1. Identify the coefficient of x, which is 'b'.
2. Take half of 'b', i.e., b/2.
3. Square this value: (b/2)².
4. Add and subtract (b/2)² to the expression: x² + bx + (b/2)² - (b/2)² + c.
5. Group the first three terms to form a perfect square: (x + b/2)².
6. Combine the constant terms: (x + b/2)² + c - (b/2)².

This results in the vertex form (x + b/2)² + k, where k = c - (b/2)².

Completing the Square for ax² + bx + c (when a ≠ 1)

When the coefficient of x², 'a', is not 1, follow these steps:

1. Factor out 'a' from the x² and x terms: a(x² + (b/a)x) + c.
2. Now, complete the square on the expression inside the parentheses, (x² + (b/a)x), using the 'b/a' term as the new 'b'. Half of (b/a) is b/(2a), and squaring it gives (b/(2a))².
3. Add and subtract this term inside the parentheses: a( (x² + (b/a)x + (b/(2a))²) - (b/(2a))² ) + c.
4. Form the perfect square: a( (x + b/(2a))² - (b/(2a))² ) + c.
5. Distribute 'a' back into the terms: a(x + b/(2a))² - a(b/(2a))² + c.
6. Simplify the constant terms: a(x + b/(2a))² + c - b²/(4a).

This gives the form a(x + b/(2a))² + k, where k = c - b²/(4a).

Using Completed Square Form to Find Vertex and Solve Equations

The completed square form, a(x - h)² + k, immediately gives the vertex of the parabola as (h, k). For a quadratic a(x + b/(2a))² + c - b²/(4a), the vertex is (-b/(2a), c - b²/(4a)). If 'a' is positive, the vertex is a minimum point; if 'a' is negative, it's a maximum point. To solve equations, set the completed square form equal to zero: a(x - h)² + k = 0. Rearrange to isolate the squared term: a(x - h)² = -k, then (x - h)² = -k/a. Take the square root of both sides: x - h = ±√(-k/a). Finally, solve for x: x = h ±√(-k/a). This method applies even when roots are irrational or complex, revealing the nature of the solutions.

Simplifying Algebraic Fractions

To simplify an algebraic fraction, you must factorise both the numerator and the denominator completely. After factorising, cancel out any common factors that appear in both the numerator and the denominator. For example, to simplify (x^2 - 4) / (x^2 + x - 2), factorise to get ((x-2)(x+2)) / ((x+2)(x-1)). The common factor (x+2) can be cancelled, leaving (x-2) / (x-1). Remember that you cannot cancel terms that are added or subtracted, only factors that are multiplied.

Multiplying and Dividing Algebraic Fractions

To multiply algebraic fractions, first factorise all numerators and denominators. Then, multiply the numerators together and the denominators together. You can cancel any common factors between any numerator and any denominator before or after multiplying. For division, 'flip' the second fraction (find its reciprocal) and change the operation to multiplication, then proceed as above. For example, (a/b) ÷ (c/d) = (a/b) × (d/c) = (ad)/(bc). Watch out for expressions like (x-y) and (y-x), where (y-x) = -(x-y).

Adding and Subtracting Algebraic Fractions

When adding or subtracting algebraic fractions, you must find a common denominator. This is usually the lowest common multiple (LCM) of the denominators. Factorising the denominators first can help identify the LCM. Once you have a common denominator, rewrite each fraction with this denominator, adjusting its numerator accordingly. Then, combine the numerators while keeping the common denominator. Finally, simplify the resulting fraction by factorising and cancelling if possible. For example, to add 1/x + 1/(x+1), the common denominator is x(x+1). This leads to (x+1)/(x(x+1)) + x/(x(x+1)) = (2x+1)/(x(x+1)).

Solving Equations with Algebraic Fractions

To solve an equation involving algebraic fractions, the first step is usually to eliminate the denominators. This is done by multiplying every term in the equation by the lowest common multiple (LCM) of all denominators. Be careful to multiply *every* term, including any terms that are not fractions. After eliminating denominators, simplify the resulting equation, which will often be a linear or quadratic equation. Solve the simplified equation using appropriate methods. Always check your solutions, as some values might make an original denominator zero, making them extraneous solutions.

Understanding Variables and Constants

An algebraic expression is a combination of variables, numbers, and arithmetic operations. A variable is a letter (like x, y, a, b) that represents an unknown number. A constant is a number whose value does not change. For example, in the expression '3x + 5', 'x' is the variable, '3' is the coefficient of x (it multiplies x), and '5' is a constant. When formulating expressions, always clearly define what each variable represents.

Translating Keywords into Operations

Many common words and phrases indicate specific mathematical operations:

• Addition (+): 'sum', 'plus', 'increased by', 'more than', 'total'
• Subtraction (-): 'difference', 'minus', 'decreased by', 'less than', 'take away'
• Multiplication (*): 'product', 'times', 'multiplied by', 'of' (e.g., 'half of x' means 0.5x)
• Division (/): 'quotient', 'divided by', 'per', 'split equally'

Be careful with subtraction and division order. For example, 'x less than 5' is 5 - x, not x - 5.

Writing Expressions Concisely

Algebraic expressions should be written in the most concise way possible. This means:

• Omit multiplication signs between numbers and variables (e.g., 3x instead of 3 * x).
• Omit multiplication signs between variables (e.g., ab instead of a * b).
• Place the number before the variable (e.g., 5y instead of y5).
• Write division as a fraction (e.g., x/4 instead of x ÷ 4).
• Use parentheses when necessary to group terms, especially when dealing with 'sum of' or 'difference of' combined with multiplication/division (e.g., 'twice the sum of x and y' is 2(x + y)).

Common Pitfalls and Examples

Avoid common mistakes such as incorrect order of operations or misinterpreting keywords. For instance, 'the difference between x and 3' is x - 3 or 3 - x, often implying the larger minus the smaller or specified order. 'Three times a number decreased by 5' is 3x - 5. However, 'three times the difference between a number and 5' would be 3(x - 5). Always read the problem carefully to ensure you capture the intended structure. Practice with varied wordings to build confidence in formulating expressions correctly.

Identifying Variables and Constants

When formulating an algebraic expression from a word problem, the first step is to clearly identify the unknown quantities. Assign distinct variables (e.g., x, y, t, A) to represent these unknowns. For example, if a problem mentions 'the number of apples' and 'the number of oranges', you might use 'A' for apples and 'O' for oranges. Constants are fixed numerical values given in the problem, such as '5 more' or 'double the amount'. Be careful to distinguish between quantities that change (variables) and quantities that remain the same (constants). Sometimes, context implies a common constant, like 'pi' in circle problems, even if not explicitly stated.

Translating Keywords into Operations

Effective formulation relies on accurate translation of verbal cues into mathematical operations.

• 'Sum', 'total', 'increased by', 'more than' indicate addition (+).
• 'Difference', 'decreased by', 'less than', 'subtracted from' indicate subtraction (-). Note the order for subtraction: '5 less than x' is x - 5.
• 'Product', 'times', 'multiplied by', 'of' (especially with fractions or percentages) indicate multiplication (× or juxtaposition, e.g., 3x).
• 'Quotient', 'divided by', 'ratio' indicate division (÷ or fraction bar).
• 'Square of', 'cubed', 'to the power of' indicate exponents.

Handling Complex Conditions and Multiple Steps

Higher-tier problems often involve multiple conditions or require combining simpler expressions. Break down the problem into sequential steps. For instance, if 'the cost of buying x items at £5 each, plus a fixed delivery charge of £3' is described, first form the expression for the cost of items (5x), then add the delivery charge (5x + 3). Use parentheses to group terms correctly, especially when a whole quantity is being multiplied, divided, or raised to a power (e.g., 'the square of x plus 3' is (x+3)²). Be alert for relationships between variables. If 'y is twice x', then y = 2x, which allows substitution to reduce the number of variables if required.

Context-Specific Expressions (e.g., Geometry, Rates)

Many formulation problems are set in specific contexts. For geometry, recall formulas for area (e.g., A = lw for a rectangle, A = πr² for a circle), perimeter, volume (e.g., V = lwh for a cuboid). Substitute algebraic expressions for known dimensions. For problems involving distance, speed, and time, remember the relationship d = st. If a problem states 'the length of a rectangle is 3 cm more than its width', let width = w, then length = w + 3. Always ensure your expression unit-wise makes sense in the context of the problem.