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Higher Tier

Quadratic Equations: Complete GCSE Guide

Learn how to solve quadratic equations using three methods: factorising, the quadratic formula, and completing the square. This is one of the most important algebra topics for GCSE Maths.

What is a Quadratic Equation?

A quadratic equation is any equation that can be written in the form:

ax² + bx + c = 0

where a ≠ 0

The key feature is the term (the "squared" term). Examples include:

Method 1: Solving by Factorising

Factorising is the quickest method when it works. You're looking for two brackets that multiply to give the original quadratic.

Steps to Factorise

Step 1: Make sure the equation equals zero
Step 2: Find two numbers that multiply to give 'c' and add to give 'b'
Step 3: Write as (x + ?)(x + ?) = 0
Step 4: Set each bracket = 0 and solve

Worked Example

Solve x² + 7x + 12 = 0

Find two numbers that multiply to give 12 and add to give 7:

→ 3 × 4 = 12 ✓ and 3 + 4 = 7 ✓

So: (x + 3)(x + 4) = 0

Either x + 3 = 0 → x = -3

Or x + 4 = 0 → x = -4

Method 2: The Quadratic Formula

The quadratic formula works for ALL quadratic equations, even when they don't factorise nicely.

x = (-b ± √(b² - 4ac)) / 2a

MEMORISE THIS! It's not given in the exam.

Worked Example

Solve 2x² + 5x - 3 = 0

Here: a = 2, b = 5, c = -3

x = (-5 ± √(25 - 4×2×(-3))) / (2×2)

x = (-5 ± √(25 + 24)) / 4

x = (-5 ± √49) / 4

x = (-5 ± 7) / 4

x = 2/4 = 0.5 or x = -12/4 = -3

💡 Exam Tip: The Discriminant

The part under the square root (b² - 4ac) is called the discriminant:

  • If b² - 4ac > 0 → Two different real solutions
  • If b² - 4ac = 0 → One repeated solution
  • If b² - 4ac < 0 → No real solutions

Method 3: Completing the Square

Completing the square rewrites ax² + bx + c in the form a(x + p)² + q. This is useful for finding the vertex of a parabola and is required for Higher tier.

Steps to Complete the Square

Step 1: Halve the coefficient of x
Step 2: Write (x + half)²
Step 3: Subtract (half)² to compensate
Step 4: Simplify the constant

Worked Example

Complete the square for x² + 6x + 5

Half of 6 = 3

= (x + 3)² - 9 + 5

= (x + 3)² - 4

To solve: (x + 3)² = 4

x + 3 = ±2

x = -1 or x = -5

Which Method Should I Use?

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