Review the key concepts, formulae, and examples before starting your quiz.
🔑Concepts
A Quadratic Equation is an algebraic equation of the second degree, typically written in the standard form , where and are real numbers and . Visually, this equation represents a parabola; solving the equation means finding where this parabola crosses the horizontal -axis.
The Roots of a quadratic equation are the values of that satisfy the equation. If we substitute a root into the equation, the result is zero. On a graph, these roots are the -intercepts of the quadratic function.
The Factorisation Method involves breaking down the quadratic expression into a product of two linear factors, such as . This is based on the idea that a large area (the quadratic) can be represented as the product of its length and width (the linear factors).
Splitting the Middle Term is the primary technique for factorisation. To factor , you must find two numbers whose sum equals and whose product equals . Once found, the middle term is replaced by these two numbers to allow for factoring by grouping.
The Zero Product Property states that if the product of two numbers or expressions is zero, then at least one of them must be zero. Mathematically, if , then either or , leading to the solutions and .
Perfect Square Trinomials are special cases where the quadratic factors into two identical linear factors, such as . Visually, this means the vertex of the parabola sits exactly on the -axis, and we say the equation has two equal real roots.
The Difference of Squares is a shortcut for equations in the form . This factors directly into , representing a parabola symmetric about the -axis with roots at and .
📐Formulae
Standard Form:
Splitting Criteria: Find such that and
Factorised Form:
Difference of Squares:
Square Identity 1:
Square Identity 2:
💡Examples
Problem 1:
Solve the quadratic equation by factorisation.
Solution:
Step 1: Identify . \nStep 2: Find two numbers that multiply to and add to . These numbers are and . \nStep 3: Split the middle term: . \nStep 4: Group the first two and last two terms: . \nStep 5: Factor out common terms: . \nStep 6: Factor out the common binomial: . \nStep 7: Apply the Zero Product Property: or . \nFinal Answer: and .
Explanation:
We use the splitting the middle term method. Since the product is positive (12) and the sum is negative (-7), both factors must be negative.
Problem 2:
Solve for : .
Solution:
Step 1: Identify . \nStep 2: Find two numbers with product and sum . These numbers are and . \nStep 3: Split the middle term: . \nStep 4: Factor by grouping: . \nStep 5: Extract the common factor: . \nStep 6: Solve for : or . \nFinal Answer: .
Explanation:
In this case, is not 1, so we must multiply and to find the target product (-12). We then find factors of -12 that sum to -1 and group the terms to solve.