Overview
What you’ll learn
- Solve quadratic equations by factorising, including rearranging into standard form first.
- Apply the quadratic formula correctly, including rounding roots and giving exact surd forms.
- Complete the square for x² + bx + c and use it to solve equations and find turning points (Higher).
- Read roots, y-intercepts and turning points from a parabola, and use its symmetry.
- Find approximate solutions to quadratic equations by reading a graph.
Specification checklist (AQA 8300 · Edexcel 1MA1 · OCR J560)
Statements are paraphrased from the shared DfE subject content and grouped by tier. Foundation content is studied by every student; Higher statements are examined on Higher-tier papers only. QuizQuip is not affiliated with AQA, Pearson Edexcel or OCR.
Foundation — every student
- 8.1Solve quadratic equations of the form x² + bx + c = 0 by factorising, including differences of two squares and equations that need rearranging first.
- 8.4Identify and interpret roots, intercepts and turning points of quadratic functions from their graphs, and use the symmetry of a parabola.
- 8.5Find approximate solutions to quadratic equations using a graph.
Higher — Higher tier only
- 8.2Solve quadratic equations using the quadratic formula x = (−b ± √(b² − 4ac))⁄2a, which must be memorised, including equations of the form ax² + bx + c = 0 and those needing rearrangement. (Higher)
- 8.3Solve quadratic equations by completing the square, and use the completed-square form to deduce the turning point of a quadratic function. (Higher)
Key points
- 1Rearrange to = 0 first, then factorise: x² − 3x = 10 becomes (x − 5)(x + 2) = 0, so x = 5 or −2.
- 2A product equals zero when a factor equals zero — both roots need stating.
- 3The quadratic formula must be memorised: x = (−b ± √(b² − 4ac))⁄2a — there is no formula sheet at GCSE.
- 4Inside the formula, quote b² − 4ac as a line: it decides how many real roots exist.
- 5Completing the square: x² + bx + c = (x + b⁄2)² + c − (b⁄2)² (Higher).
- 6The completed-square form (x − p)² + q shows the turning point (p, q) directly (Higher).
- 7A parabola is symmetric about its turning point; the roots sit equally either side of the axis of symmetry.
- 8The y-intercept of y = ax² + bx + c is always c — read it straight off.
Notes · 1 of 4
Solving by factorising
Get to zero, factorise, and set each factor to zero.
Factorising solves a quadratic because a product is zero only when a factor is zero. For x² + 5x + 6 = 0: factorise to (x + 2)(x + 3) = 0, so x + 2 = 0 or x + 3 = 0, giving x = −2 or x = −3. Both roots must be stated — one root alone loses the final mark.
Rearrange first when the equation is not in standard form: x² − 3x = 10 becomes x² − 3x − 10 = 0, which factorises as (x − 5)(x + 2) = 0, so x = 5 or x = −2. Differences of two squares solve instantly: x² − 9 = 0 gives (x + 3)(x − 3) = 0 and x = ±3.
Notes · 2 of 4
The quadratic formula
Memorise it — no GCSE paper provides it. (Higher)
Higher. Every quadratic ax² + bx + c = 0 yields to x = (−b ± √(b² − 4ac))⁄2a. For 2x² + 3x − 4 = 0: a = 2, b = 3, c = −4, so x = (−3 ± √(9 + 32))⁄4 = (−3 ± √41)⁄4. To 2 d.p. that is x = 0.85 or x = −2.35. Writing the b² − 4ac line separately keeps the sign errors away — c = −4 makes −4ac positive.
Exact answers stay in surd form when the question says so: x² − 6x + 2 = 0 gives x = (6 ± √28)⁄2 = 3 ± √7 — divide before simplifying, not after. If b² − 4ac comes out negative, the quadratic has no real roots, which is a legitimate conclusion in graph questions.
Notes · 3 of 4
Completing the square
Half of b, squared, then corrected. (Higher)
Higher. Completing the square rewrites x² + bx + c as (x + b⁄2)² + c − (b⁄2)²: take half of b, square the bracket, then correct the constant. So x² + 6x + 2 = (x + 3)² + 2 − 9 = (x + 3)² − 7. Solving (x + 3)² = 7 gives x = −3 ± √7 — the same roots the formula finds, but the form also exposes the turning point (−3, −7).
The link is the payoff: y = (x − p)² + q has its minimum at (p, q), because squares are never negative. y = x² − 4x + 1 = (x − 2)² − 3 turns at (2, −3). For y = x² − 6x + 11 = (x − 3)² + 2 the minimum is (3, 2) — positive q means the parabola never crosses the x-axis, matching the negative discriminant.
Notes · 4 of 4
Reading quadratic graphs
Roots, intercepts, symmetry and the turning point.
A parabola y = ax² + bx + c opens upwards for positive a and downwards for negative a. Its roots are where it crosses the x-axis; its y-intercept is c; its turning point sits on the axis of symmetry, halfway between the roots. For y = (x − 2)(x + 3), the roots are 2 and −3, so the axis is x = −½ and the minimum is at x = −½: y = (−2½)(2½) = −6¼, giving (−½, −6¼).
Factorised form makes roots readable; completed-square form makes the turning point readable; and graphs make approximate solutions readable — where the curve meets the x-axis is where ax² + bx + c = 0. On a worded graph question, quote roots to the accuracy the grid supports and call them approximate.
Revise
Key terms
The words examiners expect you to use precisely.
- Root
- A solution of an equation; on a graph, an x-intercept.y = (x − 2)(x + 3) has roots 2 and −3.
- Quadratic formula
- x = (−b ± √(b² − 4ac))⁄2a, memorised for the exam.Solves any quadratic.
- Discriminant
- The expression b² − 4ac under the formula’s root.Negative means no real roots.
- Completing the square
- Rewriting x² + bx + c as (x + b⁄2)² + c − (b⁄2)².x² + 6x + 2 = (x + 3)² − 7.
- Turning point
- The minimum or maximum of a parabola.y = (x − 2)² − 3 turns at (2, −3).
- Axis of symmetry
- The vertical line through the turning point.Halfway between the roots.
Show 1 more key termShow fewer key terms
- Parabola
- The U-shaped graph of a quadratic function.Opens upwards for a > 0.
Revise
Common mistakes
Where students most often lose marks in this chapter — and what to do instead.
Mistake: Solving (x − 5)(x + 2) = 0 as x = 5 only.
Both factors can be zero, so both roots count.
Instead: x = 5 or x = −2.
Mistake: Solving x² = 4x + 12 by dividing both sides by x.
Dividing by x loses the root x = 0… and here misrepresents the equation entirely.
Instead: Rearrange to x² − 4x − 12 = 0 and factorise.
Show 10 more mistakesShow fewer mistakes
Mistake: Forgetting ± in the quadratic formula.
The formula gives two roots through the plus-minus.
Instead: x = (−b + √…)⁄2a and x = (−b − √…)⁄2a.
Mistake: Substituting c = −4 as +4 inside b² − 4ac.
The sign of c travels with it.
Instead: b² − 4ac with a = 2, b = 3, c = −4 is 9 + 32 = 41.
Mistake: Halving b twice in completing the square: x² + 6x as (x + 6)².
The bracket takes half of b.
Instead: x² + 6x = (x + 3)² − 9.
Mistake: Forgetting the correction term: x² + 6x + 2 = (x + 3)² + 2.
Squaring (x + 3) adds 9 that the original lacks.
Instead: (x + 3)² + 2 − 9 = (x + 3)² − 7.
Mistake: Reading the turning point of y = (x − 2)² − 3 as (−2, −3).
The form is (x − p)² + q with turning point (p, q); the bracket sign flips.
Instead: The minimum is at (2, −3).
Mistake: Calling the y-intercept a root.
Roots sit on the x-axis; the y-intercept is where x = 0.
Instead: y = (x − 2)(x + 3) has y-intercept −6 and roots 2, −3.
Mistake: Dividing the formula by 2a only partially: (−b ± √)⁄2 then forgetting a.
The whole denominator is 2a.
Instead: x = (−3 ± √41)⁄4 when a = 2.
Mistake: Rounding roots from the graph as exact.
Grid readings are approximate.
Instead: “x ≈ 3.5 (from the graph)”.
Mistake: Assuming a negative discriminant is an error.
It correctly signals no real roots — the parabola misses the axis.
Instead: b² − 4ac < 0: state “no real roots”.
Mistake: Ignoring rearrangement: trying to factorise x² = 4x + 12 directly.
Factorising needs the standard = 0 form.
Instead: x² − 4x − 12 = 0 first.
Check yourself
Quick quiz
30 questions with instant explanations. About 5 minutes.
Question 1 of 3
- Question 1: current question
- Question 2: not answered
- Question 3: not answered
Solve a quadratic by factorising
Solve x² + 5x + 6 = 0.
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Practise
Topical questions
18 exam-style questions with model answers and mark schemes.
A set of structured and calculation questions matched to this chapter, in three tiers, each with a model answer, mark scheme and examiner tip. Higher-tier questions are tagged.
Practise topical questionsExam practice
Practice paper
A 30-mark paper in the style of a GCSE paper, with a separate mark scheme. Give yourself 40 min.
Question paper
Chapter 8 Practice Paper
30 marks · 40 min · 5 pages
10 multiple-choice questions and structured questions with answer lines. Chapter 8: solving quadratics by factorising, the quadratic formula and completing the square (Higher), roots, intercepts and turning points from graphs, symmetry, and approximate graphical solutions.
Mark scheme
PROChapter 8 Practice Paper — mark scheme
Answers · Marking points · 3 pages
Model answers with a mark for each creditworthy point — for marking at home or in class.
How is this chapter examined?
Paper 1
non-calculator · 1 h 30
Exact fractions, surds and multiples of π expected
Papers 2 and 3
calculator · 1 h 30 each
Non-exact answers to 3 s.f., angles to 1 d.p.
Tiers
Foundation (grades 1–5) · Higher (grades 4–9)
Higher assumes all Foundation content
Want a full paper? Try the Paper 1-style (Foundation), Papers 2 & 3 style (Foundation), Paper 1-style (Higher), Papers 2 & 3 style (Higher) sets.
Questions
Frequently asked questions
Do I need to memorise the quadratic formula?
Yes. GCSE Maths papers print no formula sheet, so x = (−b ± √(b² − 4ac))⁄2a must be recalled. Write it down at the start of the exam if you like — a memory dump on the first page is allowed.
When should I complete the square instead of using the formula?
When the question asks for the form (x + p)² + q, or for a turning point. Completing the square gives both the exact roots (from (x + p)² + q = 0, x = −p ± √(−q)) and the minimum point (−p, q) in one step.
Which parts are Higher only?
Solving by factorising, reading graphs and approximate solutions are common to both tiers. The quadratic formula, completing the square and turning points from the completed-square form are Higher-tier.
What does a negative discriminant tell me?
That b² − 4ac is below zero, so the square root is not real and the quadratic has no real roots — the parabola sits entirely above (or below) the x-axis. It is a conclusion, not a mistake.
Up next
Chapter 9: Simultaneous equations and inequalities