What you'll learn
Quadratic equations are a cornerstone of the CSEC Mathematics syllabus, appearing in the Algebra section and threading through Relations, Functions and Graphs. A quadratic equation is any equation that can be written in the form ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. In this guide you will learn the three main methods CSEC expects you to master — factorisation, completing the square, and the quadratic formula — and how to choose the right one under exam conditions. You will also learn to solve word problems that lead to a quadratic, to recognise when a quadratic has two solutions, one repeated solution, or none, and to connect the algebra to the shape of a parabola. Quadratics appear in Paper 1 (multiple choice) and are almost guaranteed in Paper 2, often inside a longer problem on areas, motion, or number puzzles.
Key terms and definitions
Quadratic equation — an equation of the form ax² + bx + c = 0 with a ≠ 0; the highest power of the unknown is 2.
Root (or solution) — a value of x that satisfies the equation, making the left side equal to zero.
Factorisation — writing the quadratic as a product of two linear brackets, e.g. (x + 3)(x − 2).
Completing the square — rewriting ax² + bx + c in the form a(x + p)² + q.
Quadratic formula — the formula x = [−b ± √(b² − 4ac)] ÷ 2a, which solves any quadratic.
Discriminant — the expression b² − 4ac under the square root; its sign tells you the number of real roots.
Parabola — the U-shaped (or ∩-shaped) curve that is the graph of a quadratic function.
Coefficient — the number multiplying a term; in 3x², the coefficient of x² is 3.
Core concepts
Solving by factorisation
Factorisation is the quickest method when it works. The idea rests on the zero-product principle: if two quantities multiply to give zero, at least one of them must be zero. So if (x + 3)(x − 2) = 0, then either x + 3 = 0 or x − 2 = 0, giving x = −3 or x = 2.
To factorise x² + bx + c, look for two numbers that multiply to c and add to b. For x² + x − 6, you need two numbers multiplying to −6 and adding to +1: those are +3 and −2, so it factorises to (x + 3)(x − 2). When a ≠ 1 (for example 2x² + 7x + 3), multiply a by c (2 × 3 = 6), find two numbers multiplying to 6 and adding to 7 (6 and 1), split the middle term, and factor by grouping: 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3).
Completing the square
When a quadratic does not factorise neatly, completing the square always works and is also the method behind finding the minimum or maximum point of a parabola. For x² + 6x + 5, take half the coefficient of x (half of 6 is 3), square it (9), and write x² + 6x = (x + 3)² − 9. So x² + 6x + 5 = (x + 3)² − 9 + 5 = (x + 3)² − 4. Setting this to zero: (x + 3)² = 4, so x + 3 = ±2, giving x = −1 or x = −5. The form (x + 3)² − 4 also tells you the vertex of the parabola is at (−3, −4).
The quadratic formula
The quadratic formula solves every quadratic and is the safest choice when factors are not obvious. For ax² + bx + c = 0,
x = [−b ± √(b² − 4ac)] ÷ 2a.
Carefully substitute the values of a, b and c (watching signs), simplify under the root, then evaluate both the + and the − cases. The formula is especially useful when the roots are not whole numbers, since CSEC often asks for answers correct to two or three significant figures.
The discriminant
The quantity b² − 4ac is called the discriminant. It reveals the nature of the roots before you finish solving: if b² − 4ac > 0 there are two different real roots; if b² − 4ac = 0 there is one repeated real root (the parabola just touches the x-axis); if b² − 4ac < 0 there are no real roots (the parabola does not cross the x-axis). This is a favourite short Paper 2 question.
Forming quadratics from word problems
Many CSEC problems do not arrive as a tidy equation. A rectangular garden whose length is 3 m more than its width and whose area is 40 m² gives width w and length w + 3, so w(w + 3) = 40, i.e. w² + 3w − 40 = 0. Solving gives w = 5 (rejecting the negative root, since a length cannot be negative). Always check that each root makes sense in the real context and discard impossible answers.
Worked examples
Example 1: Solving by factorisation (Paper 1/2 style)
Solve x² − 5x + 6 = 0.
Find two numbers multiplying to 6 and adding to −5: these are −2 and −3. So x² − 5x + 6 = (x − 2)(x − 3) = 0. Therefore x − 2 = 0 or x − 3 = 0, giving x = 2 or x = 3. You can check: 2² − 5(2) + 6 = 4 − 10 + 6 = 0. ✓
Example 2: Using the quadratic formula (Paper 2 style)
Solve 2x² + 3x − 4 = 0, giving answers to two decimal places.
Here a = 2, b = 3, c = −4. The discriminant is b² − 4ac = 3² − 4(2)(−4) = 9 + 32 = 41. So x = [−3 ± √41] ÷ 4. Since √41 ≈ 6.40, x = (−3 + 6.40) ÷ 4 ≈ 0.85 or x = (−3 − 6.40) ÷ 4 ≈ −2.35. The solutions are x ≈ 0.85 or x ≈ −2.35.
Example 3: A word problem (Paper 2 style)
A farmer in St Elizabeth fences a rectangular plot. The length is 2 m longer than the width, and the area is 48 m². Find the dimensions.
Let the width be w metres; the length is w + 2. Then w(w + 2) = 48, so w² + 2w − 48 = 0. Factorising: two numbers multiplying to −48 and adding to 2 are 8 and −6, giving (w + 8)(w − 6) = 0. So w = −8 or w = 6. A width cannot be negative, so w = 6 m and the length is 8 m.
Common mistakes and how to avoid them
Forgetting to set the equation to zero first. Methods rely on ax² + bx + c = 0. If you are given x² + 4x = 5, rearrange to x² + 4x − 5 = 0 before factorising.
Sign errors in the formula. The term is −b, so if b = −3 then −b = +3. Write the substitution out fully and use brackets around negative numbers: b² − 4ac = (−3)² − 4(2)(−4).
Losing a solution. A quadratic usually has two roots. After (x + 3)² = 4, remember x + 3 = +2 and x + 3 = −2. Do not take only the positive square root.
Keeping impossible answers in word problems. Lengths, ages and numbers of people cannot be negative. State that you reject the invalid root and why.
Mis-factorising when a ≠ 1. Always multiply a by c first; do not just look for factors of c.
Exam technique for Quadratic Equations
Choose the method the question signals. If it factorises easily, factorise. If it says "give your answer to 2 decimal places" or "to 3 significant figures", expect the quadratic formula. If it asks for a minimum/maximum or vertex, complete the square.
Check the discriminant if asked about the "nature of the roots". State two real, one repeated, or no real roots, and justify with the sign of b² − 4ac.
Show every step. CSEC awards method marks. Even if your final arithmetic slips, a correct factorisation or correct substitution earns marks.
Verify at least one root. Substituting back takes seconds and catches sign errors.
Read word problems twice. Define your variable clearly ("let the width be w"), form the equation, solve, then answer the actual question asked (often the dimensions, not just x).
Quick revision summary
A quadratic equation has the form ax² + bx + c = 0 with a ≠ 0 and usually has two roots. Factorisation uses the zero-product principle: find two numbers multiplying to (a × c) and adding to b, then set each bracket to zero. Completing the square rewrites the quadratic as a(x + p)² + q, which both solves the equation and locates the vertex of the parabola. The quadratic formula, x = [−b ± √(b² − 4ac)] ÷ 2a, solves any quadratic and is best when the roots are not whole numbers. The discriminant b² − 4ac tells you the nature of the roots: positive means two real roots, zero means one repeated root, negative means no real roots. For word problems, define your variable, form the equation, solve, and reject any answer that is impossible in context. Always set the equation to zero first, watch your signs, keep both roots, and show full working to secure method marks.