Quadratic Equations

GCSE Maths · Algebra

Quadratic Equations

A quadratic equation has the form ax² + bx + c = 0 where a ≠ 0. Solving a quadratic means finding the values of x that make it true. There are three main methods.

Method 1: Factorising

This is the quickest method when it works. You express the quadratic as a product of two brackets.

Worked Example: Solve x² + 5x + 6 = 0.

Find two numbers that multiply to give +6 and add to give +5: +2 and +3
(x + 2)(x + 3) = 0
Either x + 2 = 0 → x = −2, or x + 3 = 0 → x = −3

Worked Example: Solve 2x² − 7x + 3 = 0.

Multiply a × c = 2 × 3 = 6
Find two numbers that multiply to 6 and add to −7: −1 and −6
Rewrite: 2x² − x − 6x + 3 = 0
Factorise in pairs: x(2x − 1) − 3(2x − 1) = 0
(x − 3)(2x − 1) = 0
x = 3 or x = 0.5

Difference of two squares: a² − b² = (a + b)(a − b)

Worked Example: Solve x² − 49 = 0.

(x + 7)(x − 7) = 0
x = 7 or x = −7

Method 2: The Quadratic Formula

When a quadratic does not factorise neatly, use the formula:

x = (−b ± √(b² − 4ac)) / 2a

Worked Example: Solve 3x² + 2x − 4 = 0. Give answers to 2 d.p.

a = 3, b = 2, c = −4
b² − 4ac = 4 − (4 × 3 × −4) = 4 + 48 = 52
x = (−2 ± √52) / 6
x = (−2 + 7.2111) / 6 = 0.87 or x = (−2 − 7.2111) / 6 = −1.54

The Discriminant (Higher)

The expression b² − 4ac is called the discriminant. It tells you how many solutions a quadratic has:

b² − 4ac > 0 → two distinct real roots
b² − 4ac = 0 → one repeated root (the curve touches the x-axis)
b² − 4ac < 0 → no real roots (the curve does not cross the x-axis)

Method 3: Completing the Square (Higher)

Rewrite ax² + bx + c in the form a(x + p)² + q.

Method for x² + bx + c: Halve the coefficient of x, square it, then adjust.

Worked Example: Write x² + 6x + 2 in completed square form, then solve x² + 6x + 2 = 0.

Half of 6 is 3: (x + 3)² = x² + 6x + 9
But we need x² + 6x + 2, so subtract 7: (x + 3)² − 7
To solve: (x + 3)² − 7 = 0
(x + 3)² = 7
x + 3 = ±√7
x = −3 + √7 or x = −3 − √7

Why it matters: The completed square form (x + p)² + q tells you the turning point of the parabola is at (−p, q).

From the example above, the minimum point is (−3, −7).

Forming Quadratic Equations

Some problems require you to set up a quadratic from context.

Worked Example: The length of a rectangle is (x + 3) cm and the width is (x − 1) cm. The area is 45 cm². Find x.

(x + 3)(x − 1) = 45
x² + 2x − 3 = 45
x² + 2x − 48 = 0
(x + 8)(x − 6) = 0
x = −8 or x = 6
Since length must be positive, x = 6 (length = 9 cm, width = 5 cm)

Exam Tips

Always rearrange to = 0 before factorising or using the formula
When the question says "give exact answers", leave surds — do not use a calculator
If a quadratic comes from a real-world context, reject negative or nonsensical solutions
Show all steps when using the quadratic formula — marks are awarded for substitution, discriminant and final answers separately

Quadratic Equations