What you'll learn
This topic covers two powerful methods for solving quadratic equations: completing the square and the quadratic formula. In this guide you will learn how to write a quadratic in completed-square form, how to use it to solve equations and find turning points, how to apply the quadratic formula, and how to decide which method to use. These are key higher-tier algebra skills.
Key terms and definitions
Quadratic equation — an equation of the form ax² + bx + c = 0.
Completing the square — writing ax² + bx + c in the form a(x + p)² + q.
Quadratic formula — x = (−b ± √(b² − 4ac)) ÷ 2a.
Discriminant — the value b² − 4ac, which tells you the number of solutions.
Turning point — the maximum or minimum point of a parabola.
Core concepts
Completing the square
For x² + bx + c, write it as (x + b/2)² − (b/2)² + c. You halve the coefficient of x to get the number inside the bracket, then subtract its square to keep the value the same. For example, x² + 6x + 1 = (x + 3)² − 9 + 1 = (x + 3)² − 8.
Solving by completing the square
Once in completed-square form, rearrange and square root to solve. For (x + 3)² − 8 = 0: (x + 3)² = 8, so x + 3 = ±√8, giving x = −3 ± √8. This gives exact (surd) answers, useful when factorising is impossible.
Turning points from completed square
The completed square a(x + p)² + q reveals the turning point at (−p, q). For y = (x + 3)² − 8, the minimum point is (−3, −8). This is the quickest way to find the vertex of a parabola.
The quadratic formula
For any ax² + bx + c = 0, the solutions are x = (−b ± √(b² − 4ac)) ÷ 2a. Substitute a, b and c carefully, watching signs. The ± gives the two solutions. This method always works, even when factorising does not.
The discriminant
The discriminant b² − 4ac tells you how many real solutions there are: positive → two solutions, zero → one (repeated) solution, negative → no real solutions. You can use it without solving the whole equation.
Worked examples
Example 1: Completing the square
Write x² + 8x + 3 in completed-square form.
(x + 4)² − 16 + 3 = (x + 4)² − 13.
Example 2: Quadratic formula
Solve 2x² + 3x − 2 = 0.
x = (−3 ± √(9 + 16)) ÷ 4 = (−3 ± 5) ÷ 4, so x = ½ or x = −2.
Example 3: Discriminant
How many solutions does x² + 4x + 7 = 0 have?
b² − 4ac = 16 − 28 = −12 (negative), so no real solutions.
Common mistakes and how to avoid them
Forgetting to subtract (b/2)². You must subtract the square to keep the value unchanged.
Sign errors in the formula. Use brackets when substituting negative values.
Dropping the ±. A quadratic usually has two solutions.
Misreading the turning point. The x-coordinate is −p, not p.
Dividing only part by 2a. The whole numerator is divided by 2a.
Exam technique for Completing the Square and the Quadratic Formula
Halve the x-coefficient for the bracket, then subtract its square.
Use completed square for turning points and exact solutions.
Substitute carefully into the quadratic formula, using brackets.
Check the discriminant to know how many solutions to expect.
Leave answers in surd form when asked for exact values.
Quick revision summary
Completing the square rewrites x² + bx + c as (x + b/2)² − (b/2)² + c — halve the x-coefficient for the bracket, then subtract its square. This form solves equations by rearranging and square rooting (giving exact surd answers) and reveals the turning point at (−p, q) for a(x + p)² + q. The quadratic formula x = (−b ± √(b² − 4ac)) ÷ 2a solves any quadratic; substitute a, b and c carefully with brackets and keep the ± for two solutions. The discriminant b² − 4ac tells you the number of real solutions: positive gives two, zero gives one repeated, negative gives none. The common errors are forgetting to subtract (b/2)², sign slips in the formula, dropping the ±, and dividing only part of the numerator by 2a. Halve the coefficient for the square, use it for turning points and exact answers, substitute carefully into the formula, and check the discriminant.