Pearson Edexcel International GCSE in Mathematics (Specification A) · 4MA1 · Free · no account needed

Quadratic Equations

In one sentence. Four ways to solve a quadratic — factorising, the formula, completing the square, and reading a graph — why each one works, how to choose between them, and how to reason about how many solutions there are.

Algebra — one of 8 modules in this strand, and one of 53 in Pearson Edexcel International GCSE Mathematics. Both tiers.

This module ends with Muhammad al-Khwarizmi. Learn Quadratic Equations, teach it, then defend it against Muhammad al-Khwarizmi.

Where this sits in the specification

Specification 4MA1, 2.7 Quadratic equations (formula and completing the square: Higher). Incandio is aligned to this specification; it is not published by, endorsed by or affiliated with Pearson, and the sub-topic numbers are given so you can check every module against your own copy.

  • 2.7 — Quadratic equations (formula and completing the square: Higher)

1 · Understand it

What it is

A quadratic equation can be written ax² + bx + c = 0 (a ≠ 0). It has at most two real solutions (roots). There are four routes to them. FACTORISING: if (x − 7)(x + 2) = 0, one bracket must be zero, so x = 7 or x = −2. COMPLETING THE SQUARE (Higher): rewrite x² + 6x − 3 as (x + 3)² − 12, then square-root. THE FORMULA (Higher): x = (−b ± √(b² − 4ac)) / 2a works for every quadratic. A GRAPH: the solutions are where y = ax² + bx + c crosses the x-axis — which is also why there can be two, one or no real solutions.

Why it works

Factorising works because of the zero-product rule: two numbers multiply to make zero only if one of them IS zero — true of no other number, which is why the equation must equal 0 first. Completing the square works because (x + p)² = x² + 2px + p², so x² + bx is (x + b/2)² with (b/2)² taken away; once x appears only once, inside a square, you can undo the square with ±√. The formula is completing the square done once on ax² + bx + c, so it inherits both the ± (two square roots) and the b² − 4ac under the root — and that expression decides how many real solutions exist: positive, two; zero, one repeated; negative, none.

When a question is asking for it

Factorise when integer roots are likely (try it for about thirty seconds). Use the formula when the question says 'to 2 decimal places' or the numbers will not factorise. Complete the square when the question asks for exact answers in surd form, for the turning point of a graph, or to show something is always positive. Use a graph for approximate solutions or to see how many there are. In problems you will often have to BUILD the quadratic first — from an area, Pythagoras, a speed–time relation or an intersection of a line and a curve.

How to hold it in your head

Picture the parabola y = ax² + bx + c. Moving it up and down changes how many times it meets the x-axis: twice, touching once (a repeated root), or not at all. Completing the square tells you where its lowest (or highest) point is, so it tells you at a glance which of those three pictures you have. In context, each root is a candidate answer that still has to survive reality — a negative length or a time before the start is rejected, with a reason.

2 · The mistakes that cost marks

  • Solving x² + 2x − 48 = 45 by factorising the left side as it stands. The zero-product rule needs ZERO on one side — move the 45 first.
  • Giving one solution when two exist. The ± in the formula and the two brackets both produce two.
  • Formula slips: −b when b is already negative (−(−7) = +7), the −4ac term when c is negative (it becomes +), and the 2a dividing the WHOLE top, not just the root.
  • Dividing both sides by x in x² = 5x, silently discarding the solution x = 0. Factor instead: x(x − 5) = 0.
  • Completing the square on 2x² + 8x + 1 without taking the 2 out first: the method needs x² with coefficient 1, so write 2(x² + 4x) + 1 = 2[(x + 2)² − 4] + 1.
  • Forgetting the ± after square-rooting: (x + 3)² = 12 gives x + 3 = √12 OR −√12.

3 · Worked examples

Laid out the way an examiner wants working shown — every step is a mark.

1 · Factorising. Solve x² − 5x − 14 = 0.

  1. We need (x + p)(x + q) with p × q = −14 (the constant) and p + q = −5 (the x-coefficient). Why: expanding (x + p)(x + q) gives x² + (p + q)x + pq.
  2. The product is negative, so the signs differ; the sum is negative, so the larger number is negative: −7 and +2.
  3. (x − 7)(x + 2) = 0, so x − 7 = 0 or x + 2 = 0.
  4. x = 7 or x = −2. Check one: 7² − 5(7) − 14 = 49 − 35 − 14 = 0 ✓.

Answer: x = 7 or x = −2

2 · Factorising when a ≠ 1. Solve 2x² + 5x − 12 = 0.

  1. Multiply a × c: 2 × (−12) = −24. Find two numbers with product −24 and sum +5: +8 and −3.
  2. Split the middle term with them: 2x² + 8x − 3x − 12 = 0. (The value is unchanged: 8x − 3x = 5x.)
  3. Factorise in pairs: 2x(x + 4) − 3(x + 4) = 0, so (2x − 3)(x + 4) = 0.
  4. 2x − 3 = 0 gives x = 3/2 (divide by the 2 — the step most often missed); x + 4 = 0 gives x = −4.

Answer: x = 1.5 or x = −4

3 · The quadratic formula. Solve 2x² − 3x − 4 = 0, giving your answers to 2 decimal places.

  1. 'To 2 decimal places' signals the roots are not whole numbers, so go straight to the formula. Identify a = 2, b = −3, c = −4 — signs included.
  2. Discriminant first, with brackets: b² − 4ac = (−3)² − 4(2)(−4) = 9 + 32 = 41. Positive, so there are two solutions.
  3. x = (−b ± √41) / 2a = (3 ± √41) / 4. Note −b = −(−3) = +3, and the 4 divides the WHOLE top.
  4. √41 = 6.4031…, so x = (3 + 6.4031)/4 = 2.3508… or x = (3 − 6.4031)/4 = −0.8508…
  5. Check by substituting the unrounded value 2.3508 into 2x² − 3x − 4: 11.052 − 7.052 − 4 ≈ 0 ✓. Round only at the end.

Answer: x = 2.35 or x = −0.85

4 · Completing the square (exact answers). Solve x² + 6x − 3 = 0, giving your answers in surd form.

  1. Halve the x-coefficient: 6 ÷ 2 = 3. Because (x + 3)² = x² + 6x + 9, the expression x² + 6x is (x + 3)² − 9.
  2. So x² + 6x − 3 = (x + 3)² − 9 − 3 = (x + 3)² − 12. Now x appears only once, inside the square.
  3. (x + 3)² − 12 = 0 → (x + 3)² = 12 → x + 3 = ±√12. The ± is essential: (−√12)² is also 12.
  4. √12 = √4 × √3 = 2√3, so x = −3 ± 2√3.

Answer: x = −3 + 2√3 or x = −3 − 2√3

5 · How many solutions? Show that 2x² − 4x + 5 = 0 has no real solutions.

  1. Method A — complete the square. Take out the 2: 2(x² − 2x) + 5 = 2[(x − 1)² − 1] + 5 = 2(x − 1)² + 3.
  2. (x − 1)² is never negative, so 2(x − 1)² + 3 is always at least 3. It can never equal 0, so there are no real solutions.
  3. Method B — the discriminant: b² − 4ac = (−4)² − 4(2)(5) = 16 − 40 = −24 < 0. The formula would need the square root of a negative number, so there are no real roots.
  4. Graph meaning: the lowest point of y = 2x² − 4x + 5 is (1, 3), above the x-axis, so the curve never meets it. A discriminant of 0 would mean the curve just touches the axis (one repeated root).

Answer: No real solutions: 2(x − 1)² + 3 ≥ 3 > 0 for every x

6 · Roots, turning point and graph. For y = x² − 4x − 5, find where the curve crosses the axes and its turning point.

  1. x-axis: solve x² − 4x − 5 = 0 → (x − 5)(x + 1) = 0 → x = 5 or x = −1. Crossing points (5, 0) and (−1, 0).
  2. y-axis: put x = 0 → y = −5. Crossing point (0, −5).
  3. Complete the square: x² − 4x − 5 = (x − 2)² − 4 − 5 = (x − 2)² − 9. The square is smallest (zero) when x = 2, so the minimum point is (2, −9).
  4. Check the symmetry: the line of symmetry x = 2 is exactly halfway between the roots −1 and 5 — it always is.

Answer: Roots (−1, 0) and (5, 0); y-intercept (0, −5); turning point (2, −9)

7 · A line meeting a curve (Higher). Find where y = 2x − 3 meets y = x² − 3x + 1.

  1. Where they meet, the y-values are equal: x² − 3x + 1 = 2x − 3.
  2. Rearrange to equal zero: x² − 5x + 4 = 0 → (x − 1)(x − 4) = 0 → x = 1 or x = 4.
  3. Substitute each x into the LINE (simpler) for y: x = 1 gives y = −1; x = 4 gives y = 5.
  4. Two roots means two meeting points. Had the quadratic had a repeated root, the line would be a tangent; no real roots, the line would miss the curve.

Answer: (1, −1) and (4, 5)

8 · Building the quadratic in context. A rectangle has length (x + 3) cm, width (x − 1) cm and area 45 cm². Find x.

  1. Area = length × width: (x + 3)(x − 1) = 45.
  2. Expand and move everything to one side: x² + 2x − 3 = 45 → x² + 2x − 48 = 0.
  3. Product −48, sum +2: +8 and −6, so (x + 8)(x − 6) = 0 and x = 6 or x = −8.
  4. Test both against reality: x = −8 gives a width of −9 cm, impossible — reject it and say why. x = 6 gives 9 cm × 5 cm = 45 cm² ✓.

Answer: x = 6

4 · Prove it — the five questions

The five questions climb Grade 6 → 9+ and are marked inside Incandio on your own device, against authored accepted forms, with a diagnosis of the wrong answer you actually gave. The answers stay in the app so that the practice is worth doing; the questions themselves are here.

  1. Grade 6 — Solve x² − 3x − 18 = 0. Give the positive solution.
  2. Grade 7 — Solve x² + 3x − 10 = 0. Give the negative solution.
  3. Grade 8 — Solve 2x² − 7x + 3 = 0. Give the smaller solution.
  4. Grade 9 — Solve x² − 4x − 3 = 0 using the quadratic formula. Give the larger solution correct to 3 significant figures.
  5. 9+ — The equation x² − 2kx + 3k + 4 = 0, where k is a positive constant, has exactly one solution. Work out the value of k.

Think you understand it?

Reading is the start. On Incandio a topic counts as mastered only once you have taught it and defended it — the two things a revision guide cannot ask of you.

Incandio this topic → Prove it. Teach it. Defend it. What does “Incandio it” mean?

  1. Prove it — five questions on Quadratic Equations, marked on the spot, that tell you what was missing rather than just what was wrong.
  2. Teach it — Explain how Quadratic Equations works — and why — to Ember, an AI pupil who does not have it yet. The method matters more than the vocabulary.
  3. Defend it — Argue Quadratic Equations against Muhammad al-Khwarizmi — three rounds, marked against descriptors you can read before you begin.
  4. Then the exam — read all 10 authored exam-style questions on this unit, each with its command word and its tariff, then have them marked on Incandio with the mistake you actually made named.

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