Pearson Edexcel International GCSE Mathematics A 4MA1: Quadratic equations – Revision Notes

These notes condense section 2.7, Quadratic equations, of the Pearson Edexcel International GCSE Mathematics A (4MA1) specification, Issue 2 (November 2017), for the January and June series examined on that specification. Foundation tier 2.7 is solving x² + bx + c = 0 by factorisation only. The rest – factorising ax² + bx + c, the quadratic formula, completing the square, forming equations from a context and linear-quadratic simultaneous equations – is Higher tier only. A calculator may be used on every paper.

For full explanations and worked examples, use the quadratic equations study guide. For exam-style questions with mark schemes, use the quadratic equations practice questions. Factorising and completing the square as expressions (2.2) are in the algebraic manipulation revision notes. The course hub and the printable checklist cover the rest of 4MA1, and the free diagnostics show which topics need most work.

Definitions

• Quadratic equation: an equation that can be written as ax² + bx + c = 0 with a ≠ 0.
• Solution (root): a value of x that makes the equation true. A quadratic has at most two.
• Factorised form: (x − p)(x − q) = 0, with solutions x = p and x = q.
• Completed-square form: (x + p)² + q = 0 (Higher tier only).
• Exact answer: a surd or fraction left unrounded, e.g. 4 ± √3 or 5/3.

The one rule behind factorising

If A × B = 0, then A = 0 or B = 0.

It only works with zero on one side. Always rearrange first.

Formulas and facts

What	Result	Tier
Quadratic formula	x = (−b ± √(b² − 4ac)) / 2a	Higher tier only (on the Higher formulae sheet)
Completing the square	x² + bx = (x + b/2)² − (b/2)²	Higher tier only
Difference of two squares	x² − k² = (x − k)(x + k)	Both (x² coefficient 1 on Foundation)
b² − 4ac > 0	two different solutions	Higher tier only
b² − 4ac = 0	two equal solutions	Higher tier only
b² − 4ac < 0	no real solutions	Higher tier only

Method in steps

Factorisation, x² + bx + c = 0 (both tiers)

1. Rearrange so the right-hand side is 0.
2. Find two numbers: product = c, sum = b.
3. Write (x + m)(x + n) = 0.
4. Set each bracket = 0 and solve.

Reminder: x² + 11x + 28 = 0 → (x + 4)(x + 7) = 0 → x = −4 or x = −7.

Factorisation, ax² + bx + c = 0 (Higher tier only)

1. Find two numbers: product = a x c, sum = b.
2. Split bx using those two numbers.
3. Factorise in pairs, then take out the common bracket.
4. Solve each bracket: (px + q) = 0 gives x = -q/p.

Reminder: 5x² − 13x − 6 = 0. a × c = −30, and −15 + 2 = −13. So 5x² − 15x + 2x − 6 = 5x(x − 3) + 2(x − 3) = (5x + 2)(x − 3) = 0, giving x = −2/5 or x = 3.

Quadratic formula (Higher tier only)

1. Rearrange to ax^2 + bx + c = 0.
2. Write a = , b = , c =  with their signs.
3. Work out b^2 - 4ac on its own first.
4. Substitute; write the exact form.
5. Evaluate both values; round only at the end.

Reminder: x² + 6x + 2 = 0. a = 1, b = 6, c = 2, so b² − 4ac = 36 − 8 = 28. Then x = (−6 ± √28)/2, which is x = −0.354 or x = −5.65 to 3 significant figures. The exact form simplifies to −3 ± √7.

Completing the square (Higher tier only)

1. If a is not 1, factor a out of the x terms.
2. Halve the coefficient of x: x^2 + bx -> (x + b/2)^2 - (b/2)^2.
3. Tidy to (x + p)^2 = k.
4. Square-root both sides: x + p = ±√k.
5. x = -p ± √k.

Reminder with a ≠ 1: 3x² + 12x − 5 = 0. Factor out 3: 3(x² + 4x) − 5 = 0, so 3[(x + 2)² − 4] − 5 = 0 and 3(x + 2)² − 17 = 0. Then (x + 2)² = 17/3 and x = −2 ± √(17/3), which is 0.380 or −4.38 to 3 significant figures. The −4 inside the square bracket must be multiplied by 3 before you combine it with −5.

Forming an equation from a context (Higher tier only)

1. Write the equation from the facts (area, Pythagoras, time = distance / speed).
2. Rearrange to = 0 and solve.
3. Reject impossible values and say why.
4. Answer the actual question, with units.

Reminder: two consecutive positive integers have a product of 132. Let the smaller be n, so n(n + 1) = 132 and n² + n − 132 = 0. This factorises as (n + 12)(n − 11) = 0. The integers must be positive, so n = 11 and the integers are 11 and 12.

Which method?

Clue in the question	Method
Small whole-number coefficients, no rounding asked for	Try factorising first
“Give your answers correct to 3 significant figures”	Quadratic formula
“Give your answers in the form a ± √b”	Completing the square (or formula, then simplify)
“Hence” after “write in the form (x + p)² + q”	Completing the square, using that part
Just x² and a number, e.g. 2x² = 50	Rearrange to x² = k, then ±√k
A line and a curve	Substitute, then any of the above

Linear and quadratic simultaneous equations (Higher tier only)

1. Make x or y the subject of the LINEAR equation.
2. Substitute into the quadratic; expand brackets fully.
3. Solve the resulting quadratic.
4. Find each partner value from the LINEAR equation.
5. Give the answers as pairs.

Reminder: y = 3x and x² + y² = 40 → x² + 9x² = 40 → x² = 4 → x = ±2. Pairs: (2, 6) and (−2, −6).

Must-know distinctions

• Expression vs equation. “Factorise x² − 9x + 20” wants (x − 4)(x − 5). “Solve x² − 9x + 20 = 0” wants x = 4 or x = 5. Giving one when asked for the other loses the final mark.
• Exact vs rounded. “Give exact answers” or “in the form a ± √b” means leave the surd. “Correct to 3 significant figures” means evaluate.
• (2x − 5) = 0 vs (x − 5) = 0. The first gives x = 5/2, the second x = 5.
• x² = 7x vs x = 7. Dividing by x loses x = 0. Factorise instead: x(x − 7) = 0.
• Line meets curve. Each solution pair is a point where the line meets the curve. Two pairs means two crossing points.
• “Hence”. Use the previous part. A correct answer found another way may get no credit.

Quick self-test

1. Solve x² + 2x − 24 = 0.
2. Solve x² − 49 = 0.
3. Solve x² + 5x = 0.
4. (Higher) Solve 2x² − 7x + 3 = 0.
5. (Higher) Solve 4x² = 9.
6. (Higher) Solve x² − 4x − 7 = 0 by completing the square. Give exact answers.
7. (Higher) Solve 5x² + 2x − 4 = 0. Give your answers to 3 significant figures.
8. (Higher) Work out b² − 4ac for x² + 3x + 5 = 0. What does it tell you?
9. (Higher) Solve the simultaneous equations y = x + 4 and y = x² − 2.
10. (Higher) A square has side x cm. Its area, in cm², is 21 more than its perimeter, in cm. Form an equation and find x.

Answers

1. (x + 6)(x − 4) = 0, so x = −6 or x = 4.
2. (x − 7)(x + 7) = 0, so x = 7 or x = −7.
3. x(x + 5) = 0, so x = 0 or x = −5.
4. a × c = 6; −6 and −1 add to −7. (2x − 1)(x − 3) = 0, so x = 1/2 or x = 3.
5. x² = 9/4, so x = 3/2 or x = −3/2. Or (2x − 3)(2x + 3) = 0.
6. (x − 2)² − 4 − 7 = 0, so (x − 2)² = 11 and x = 2 ± √11.
7. b² − 4ac = 4 + 80 = 84. x = (−2 ± √84)/10, so x = 0.717 or x = −1.12.
8. 3² − 4(1)(5) = 9 − 20 = −11. It is negative, so the equation has no real solutions.
9. x² − 2 = x + 4, so x² − x − 6 = 0 and (x − 3)(x + 2) = 0. x = 3, y = 7 and x = −2, y = 2.
10. x² = 4x + 21, so x² − 4x − 21 = 0 and (x − 7)(x + 3) = 0. x = −3 is impossible for a length, so x = 7. Check: area 49, perimeter 28, difference 21.

Where marks are usually lost

• Writing (x + 3)(x − 6) = 4 and then “x + 3 = 4”, instead of expanding and rearranging to zero first.
• Factorising correctly and stopping, when the question said “solve”.
• Taking the wrong sign from a bracket: (x + 6) = 0 gives x = −6, not 6.
• Dividing by x, so x = 0 disappears from x² = kx.
• Dropping the ± when square-rooting, which loses the second solution.
• In the formula, writing −b as b when b is already negative, or dividing only the square root by 2a.
• Rounding b² − 4ac or the square root early, so the final answer is wrong in the third significant figure.
• In a context question, leaving a negative length or time in the final answer, or giving x when the question asked for a length.
• In simultaneous equations, expanding (x + 2)² as x² + 4, or matching each x with the wrong y.
• Ignoring “hence” and using a different method, which can lose the method marks.

Official syllabus

Pearson Edexcel International GCSE in Mathematics (Specification A) (4MA1), Specification Issue 2, November 2017, first examination June 2018, published by Pearson Education Limited. Section 2.7 Quadratic equations (Foundation tier statement A; Higher tier statements A to D).