Quadratic Equations: GCSE Maths Methods Explained

A quadratic equation has the form ax² + bx + c = 0. Three methods solve them at GCSE: factorisation (quickest when it works), the quadratic formula (always works), and completing the square (versatile and extends to other skills). Choosing the right method and executing it accurately is what the exam tests.

How do you rearrange a quadratic equation into standard form?

Before applying any method, the equation must be in the form ax² + bx + c = 0, with all terms on one side and zero on the other.

Worked example: Rearrange x² + 3x = 10 into standard form.

Subtract 10 from both sides: x² + 3x − 10 = 0.

The equation is now in standard form with a = 1, b = 3, c = −10.

A second common situation is an equation where the quadratic is on the right-hand side:

Worked example: Rearrange 5 = x² − 2x.

Subtract 5 from both sides: 0 = x² − 2x − 5, or equivalently x² − 2x − 5 = 0.

Standard form: a = 1, b = −2, c = −5.

Always rearrange fully before choosing a solution method. Never try to factorise or apply the formula to a non-standard form.

How do you solve a quadratic by factorisation?

Factorisation is the fastest method when the quadratic factors neatly. It works by writing the equation as a product of two brackets, then using the fact that if A × B = 0 then A = 0 or B = 0.

Worked example 1: Solve x² + 5x + 6 = 0.

Find two numbers that multiply to c = 6 and add to b = 5. These are 2 and 3.

Factorise: (x + 2)(x + 3) = 0.

Set each factor to zero: x + 2 = 0 → x = −2; x + 3 = 0 → x = −3.

Solutions: x = −2 or x = −3.

Verify: (−2)² + 5(−2) + 6 = 4 − 10 + 6 = 0 ✓; (−3)² + 5(−3) + 6 = 9 − 15 + 6 = 0 ✓

Worked example 2: Solve x² − 3x = 10.

Rearrange: x² − 3x − 10 = 0.

Find two numbers that multiply to −10 and add to −3. These are −5 and 2.

Factorise: (x − 5)(x + 2) = 0.

Solutions: x = 5 or x = −2.

When a ≠ 1, try factorising by inspection or use the factor method (also called the AC method) after multiplying a and c together.

How do you use the quadratic formula?

The quadratic formula works for any quadratic equation in standard form, including those that do not factorise. It is:

x = (−b ± √(b² − 4ac)) / 2a

The expression under the square root, b² − 4ac, is called the discriminant. It tells you how many solutions exist:

Discriminant > 0: two distinct real solutions

Discriminant = 0: one repeated solution

Discriminant < 0: no real solutions

Worked example 3: Solve 2x² + 5x − 3 = 0 using the quadratic formula.

Identify: a = 2, b = 5, c = −3.

Calculate the discriminant: b² − 4ac = 25 − 4(2)(−3) = 25 + 24 = 49.

Apply the formula: x = (−5 ± √49) / 4 = (−5 ± 7) / 4.

Two solutions: x = (−5 + 7)/4 = 2/4 = 1/2 and x = (−5 − 7)/4 = −12/4 = −3.

Verify: 2(1/2)² + 5(1/2) − 3 = 1/2 + 5/2 − 3 = 3 − 3 = 0 ✓

When the question asks for answers to a given number of decimal places, this signals that the quadratic does not factorise — go straight to the formula.

How do you solve a quadratic by completing the square?

Completing the square rewrites ax² + bx + c in the form a(x + p)² + q, from which x is isolated by inverse operations. For a = 1:

Halve the coefficient of x to get p: p = b/2.

Write (x + p)² − p² + c = 0.

Rearrange to find x.

Worked example 4: Solve x² + 6x + 5 = 0 by completing the square.

Halve the coefficient of x: 6 ÷ 2 = 3.

Write: (x + 3)² − 9 + 5 = 0 → (x + 3)² − 4 = 0.

Add 4 to both sides: (x + 3)² = 4.

Square root both sides: x + 3 = ±2.

Solutions: x = −3 + 2 = −1 and x = −3 − 2 = −5.

Verify: (−1)² + 6(−1) + 5 = 1 − 6 + 5 = 0 ✓; (−5)² + 6(−5) + 5 = 25 − 30 + 5 = 0 ✓

How do you choose the right method?

Situation | Best method
Equation looks like it factorises (integer coefficients, small values) | Factorisation
Equation does not factorise, or decimal/surd answers expected | Quadratic formula
Question asks "give answer to 2 d.p." | Quadratic formula
Question asks for completing the square explicitly | Completing the square
Need the vertex form or minimum/maximum point | Completing the square

In an exam, if factorisation does not work after a minute of trying, switch to the formula — it always works and loses no marks.

What are the most common mistakes?

Mistake | Example | Fix
Not rearranging to = 0 first | Factorising x² + 3x = 10 as (x)(x + 3) = 10 | Always move all terms to one side first
Sign error in the formula | Writing −b as +b | Copy a, b, c carefully; write them out before substituting
Forgetting the ± symbol | Getting only one solution from the formula | The ± gives two solutions — always write both
Incorrect sign in factorisation | (x + 5)(x − 2) = 0 → x = 5 or x = 2 (both wrong sign) | Solving x + 5 = 0 gives x = −5, not x = 5

Frequently asked questions

What is the discriminant and what does it tell you?

The discriminant is b² − 4ac, the expression under the square root in the quadratic formula. It tells you how many real solutions the equation has without your needing to solve it fully. A positive discriminant means two distinct solutions; zero means one repeated solution (a "double root"); negative means no real solutions. At Higher GCSE this is tested explicitly — for example, "show that the equation has no real solutions" means demonstrating the discriminant is negative.

Can a quadratic equation have more than two solutions?

No. A quadratic equation has at most two real solutions. This is because a quadratic curve (parabola) can cross the x-axis at most twice. If a calculation produces three or more values of x, an error has occurred somewhere in the working. The discriminant determines whether there are two distinct solutions, one repeated solution, or no real solutions at all.

How do you solve a quadratic when there is no constant term (c = 0)?

When c = 0, the equation is ax² + bx = 0, which factorises by taking out x as a common factor: x(ax + b) = 0. This gives x = 0 or x = −b/a. Do not divide both sides by x at the start — dividing by x loses the solution x = 0. For example, x² + 4x = 0 factorises as x(x + 4) = 0, giving x = 0 or x = −4.

What does it mean for a quadratic to have repeated roots?

A repeated root occurs when the discriminant equals zero, so the square root in the formula gives ±0. Both solutions are the same value: x = −b / (2a). Geometrically this means the parabola just touches the x-axis at one point (a tangent point) rather than crossing it. For example, x² − 6x + 9 = 0 factorises as (x − 3)² = 0, giving the repeated root x = 3.

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