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A Algebra · sequences and functions — notes on one page

Every Edexcel GCSE Maths 1MA1 specification point in A, read straight through: 8 points with definitions, explanations, mark-scheme wording and worked examples. Practice questions are in the printable question pack.

Checked against Edexcel 1MA1 section A. Review basis: the qualification registry sourced from the Pearson Edexcel Level 1/Level 2 GCSE (9-1) in Mathematics (1MA1) specification; registry verification recorded 9 July 2026.

How this checking works

A18 · Solve quadratic equations (including those requiring rearrangement) algebraically by factorising, by completing the square and by using the quadratic formula; find approximate solutions using a graph

Definition. Specification A18: solve quadratic equations, including those that require rearrangement, algebraically by factorising, by completing the square and by using the quadratic formula; find approximate solutions using a graph.

  • Rearrange to ax2+bx+c=0ax^2+bx+c=0 first; expanding or clearing a fraction is often the first mark. Factorise when integer factors exist, then set each bracket to zero to get both roots.
  • Higher tier: completing the square writes x2−8x+3x^2-8x+3 as (x−4)2−13(x-4)^2-13, so (x−4)2=13(x-4)^2=13 and x=4±13x=4\pm\sqrt{13}; with a coefficient of x2x^2 above 11, take it out as a factor first.
  • The quadratic formula x=−b±b2−4ac2ax=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} works for any quadratic; write the substitution line with the signs shown, because sign errors are what the reports blame.
  • Give both solutions unless the context rules one out, in exact surd form when asked and otherwise rounded only at the end.
  • From a graph, the solutions of f(x)=0f(x)=0 are the xx-intercepts, and f(x)=kf(x)=k is read where the curve meets the line y=ky=k.

Worked example

1

Solve x2−8x+3=0x^2-8x+3=0. Give your answers in the form a±ba\pm\sqrt{b}, where aa and bb are integers.

(3)

(Total for Question 1 is 3 marks)

Working

  1. 1.Complete the square: x2−8x+3=(x−4)2−16+3=(x−4)2−13x^2-8x+3=(x-4)^2-16+3=(x-4)^2-13.
  2. 2.Set it equal to zero and isolate the square: (x−4)2=13(x-4)^2=13.
  3. 3.Take both square roots: x−4=±13x-4=\pm\sqrt{13}.
  4. 4.Add 44 to each: x=4±13x=4\pm\sqrt{13}.

Answer: x=4±13x=4\pm\sqrt{13}

Mark scheme

Mark scheme for question 1
QuestionAnswerMarkAdditional guidance
1for (x−4)2−13=0(x-4)^2-13=0 or (x−4)2−16+3(x-4)^2-16+3, or for correct substitution into the quadratic formula, eg 8±64−122\dfrac{8\pm\sqrt{64-12}}{2}M1
for x−4=±13x-4=\pm\sqrt{13} or 8±522\dfrac{8\pm\sqrt{52}}{2}M1oe
for 4±134\pm\sqrt{13}A1cao; both solutions required

Mark-scheme wording to use

  • “M1 For (x − 3)^2 – 9 – 8 (= 0) or (x =) (−(−6) ± √((−6)^2 − 4(1)(−8)))/(2(1)) allow sign error for b M1 For x – 3 = ± √17 or x = (6 ± √68)/2 A1 cao” — Pearson 1MA1, Specimen 2 Paper 1H, Q17
  • “M1 Substitute into quadratic formula - allow sign errors M1 Evaluate as far as (5 ± √13)/2” — Pearson 1MA1, Sample Paper 3H, Q11
  • “M1 for factorisation eg (x ± 3)(x ± 5) or (x + a)(x + b) where ab = – 15 or a + b = – 2 or for substitution into quadratic formula … M1 for (x + 3)(x – 5) or for (2 ± √64)/2 A1 cao” — Pearson 1MA1, November 2024 Paper 1F, Q26

What the examiner wants. Signs under control. The June 2024 Paper 2H report says completing the square with a coefficient of x2x^2 greater than 11 is always a struggle, with credit for taking out the factor of 22 but very few completing it, and that where the quadratic formula was used it was invariably sign errors that led to incorrect solutions.

Common mistakes

  • Don't take only the positive square root and lose one solution.
  • Don't substitute bb into the formula without its sign, so −b-b is evaluated incorrectly.
  • Don't round a surd before the final answer and lose accuracy.

Exam tip. Write the substitution into the quadratic formula before evaluating; this is normally where the first method mark is earned.

A19 · Solve two simultaneous equations in two variables (linear/linear or linear/quadratic) algebraically; find approximate solutions using a graph

  • A simultaneous solution is an ordered pair satisfying both equations; graphically it is an intersection point. For two linear equations, make one variable's coefficients equal or opposite, then add or subtract to eliminate it.
  • Substitute the first value back to find the second.
  • At Higher tier, when a line and a quadratic are solved together, substitute the linear expression into the quadratic, solve it, and find the matching second coordinate for every root.
  • The examiner expects complete ordered pairs and a check in both original equations.
  • Two xx-values alone are incomplete because each belongs to a different point.

Worked example

Solve simultaneously 2x+y=112x+y=11 and x−y=1x-y=1.

  1. 1.Add the equations to eliminate yy: 3x=123x=12.
  2. 2.Solve to get x=4x=4.
  3. 3.Substitute into x−y=1x-y=1: 4−y=14-y=1, so y=3y=3.

Answer: (x,y)=(4,3)(x,y)=(4,3).

Common mistakes

  • Don't add equations whose variable coefficients are not equal or opposite, so no variable is eliminated.
  • Don't find two possible xx-values in a linear-quadratic pair but give no corresponding yy-values.

Exam tip. Write each solution as an ordered pair and substitute both coordinates into both equations when the tariff allows a checking mark.

A20 · Find approximate solutions to equations numerically using iteration [Higher only]

  • Higher tier only. Iteration rewrites an equation as x=g(x)x=g(x) and repeatedly applies xn+1=g(xn)x_{n+1}=g(x_n) from the stated starting value.
  • Each new output becomes the next input. Keep the calculator's full stored value and round only values you are asked to record, because premature rounding can change later iterates.
  • Convergence is indicated when successive values agree to the required accuracy, although not every rearrangement converges.
  • The examiner expects the requested named iterate or a justified approximate root, with enough intermediate values to show the process.
  • Substitute the final approximation into the original equation as a reasonableness check.

Worked example

The iteration xn+1=10−xnx_{n+1}=\sqrt{10-x_n} starts with x0=3x_0=3. Find x1x_1 and x2x_2 to three decimal places.

  1. 1.x1=10−3=7=2.64575…x_1=\sqrt{10-3}=\sqrt7=2.64575\ldots.
  2. 2.Use the unrounded value: x2=10−2.64575…=2.71187…x_2=\sqrt{10-2.64575\ldots}=2.71187\ldots.
  3. 3.Round each requested result to three decimal places.

Answer: x1=2.646x_1=2.646 and x2=2.712x_2=2.712.

Common mistakes

  • Don't substitute x0x_0 again when calculating x2x_2 instead of using x1x_1.
  • Don't round each iterate heavily before using it as the next input.

Exam tip. Use the calculator answer key to carry the full previous iterate into the next substitution.

A21 · Translate simple situations or procedures into algebraic expressions or formulae; derive an equation (or two simultaneous equations), solve the equation(s) and interpret the solution

Definition. Specification A21: translate simple situations or procedures into algebraic expressions or formulae; derive an equation (or two simultaneous equations), solve the equation(s) and interpret the solution.

  • Start by defining the unknown in words, with units: let xx cm be the width. Write every other quantity as an expression in xx from the information given, such as 2x2x, 2x−62x-6 or x+3x+3.
  • The equation comes from the one fact that links them:
  • a perimeter, an angle sum of 90∘90^\circ, 180∘180^\circ or 360∘360^\circ, a total number of counters, or two expressions for equal sides. Solve it by the usual balancing steps, then answer the question that was asked, which is rarely the value of xx itself: substitute back to find the angle, the side length, the area or the ratio.
  • Schemes give a process mark for the expressions, one for the equation, one for solving and one for using the solution, so each stage must be visible.
  • A root that gives a negative length or a non-integer count is rejected with a reason. Label the final answer with what it represents.

Worked example

1

A rectangle has width xx cm. Its length is 33 cm more than its width. The perimeter of the rectangle is 3434 cm. Work out the length of the rectangle.

(3)

(Total for Question 1 is 3 marks)

Working

  1. 1.Write the length as an expression in xx: (x+3)(x+3) cm.
  2. 2.Form the perimeter equation: 2x+2(x+3)=342x+2(x+3)=34.
  3. 3.Expand and collect: 4x+6=344x+6=34, so 4x=284x=28 and x=7x=7.
  4. 4.Answer the question asked: the length is x+3=10x+3=10 cm.

Answer: 1010 cm

Mark scheme

Mark scheme for question 1
QuestionAnswerMarkAdditional guidance
1for forming a correct equation, eg 2x+2(x+3)=342x+2(x+3)=34 or x+x+3+x+x+3=34x+x+3+x+x+3=34P1oe
(dep) for a process to solve the equation, eg 4x=34−64x=34-6 or x=7x=7P1
for 1010A1cao

Mark-scheme wording to use

  • “P1 for process to develop 3 algebraic expressions, eg. (R =) n, (S =) 2n, (T =) 2n – 6, oe, at least two must be correct. … P1 for process to sum 3 algebraic expressions and equating to 54, eg. n + “2n” + “2n – 6” = 54 … P1 for start of process to solve the correct linear equation, eg. 5n = 54 + 6 (n = 12) or for 12, 24, 18 P1 for “12” : 2 × “12” – 6 oe eg 12 : 18 oe or 18 : 12 linked to T, R A1 for 1.5 or 3/2 or 1 1/2” — Pearson 1MA1, June 2022 Paper 3H, Q3
  • “P1 process to start problem solving eg forms an appropriate equation P1 complete process to solve their equation A1 cao” — Pearson 1MA1, Specimen 1 Paper 1H, Q1
  • “for process to use the perimeter of ABCD, eg 2x + 2× “4.5” = 26” — Pearson 1MA1, June 2019 Paper 1H, Q7

What the examiner wants. An equation before anything else. The November 2024 Paper 2H report says the first mark was for forming an appropriate equation from the diagram, either summing all the sides to 7272 or equating the two equal sides, the second for the process to isolate the terms in xx, and that most who reached that point solved correctly.

Common mistakes

  • Don't form x+(x+3)=34x+(x+3)=34 and count only half of the rectangle's perimeter.
  • Don't stop at x=7x=7 without finding and labelling both requested dimensions.
  • Don't keep a negative algebraic root even though the quantity is a length or age.

Exam tip. Define the variable before forming the equation and finish with a sentence interpreting every required quantity.

A22 · Solve linear inequalities in one or two variable(s), and quadratic inequalities in one variable; represent the solution set on a number line, using set notation and on a graph

  • Solve a linear inequality like an equation, reversing the inequality sign only when multiplying or dividing by a negative number.
  • On a number line, use a filled endpoint when equality is included and an open endpoint for a strict inequality.
  • At Higher tier, represent two-variable inequalities by drawing each boundary and testing a point to choose the region; strict boundaries are dashed.
  • For a quadratic inequality, find the roots and test the intervals they define, because the required values may lie inside or outside the roots.
  • The examiner expects the correct endpoint style, shading and notation as well as the algebraic boundary values.
A strict inequality outside two critical values uses open endpoints and outward shading.

Worked example

Higher tier: Solve (2x+3)(x−2)>0(2x+3)(x-2)>0.

  1. 1.Find the critical values: 2x+3=02x+3=0 gives x=−32x=-\dfrac32, and x−2=0x-2=0 gives x=2x=2.
  2. 2.Test the three intervals; the product is positive outside the roots.
  3. 3.The inequality is strict, so neither endpoint is included.

Answer: x<−32x<-\dfrac32 or x>2x>2.

Common mistakes

  • Don't forget to reverse the inequality sign after dividing by a negative number.
  • Don't use filled endpoints for << or >>.
  • Don't assume a quadratic inequality is always satisfied between its roots without testing signs.

Exam tip. For a graphical answer, the boundary style and the direction of shading are separate marking points.

A23 · Generate terms of a sequence from either a term-to-term or a position-to-term rule

  • A term-to-term rule produces each new term from the preceding term or terms, so begin with every stated starting value and apply the operations in the given order.
  • A position-to-term rule gives a term directly from its position nn; for the first terms substitute n=1,2,3,…n=1,2,3,\ldots unless the question explicitly defines another starting index.
  • Multi-step, alternating and Fibonacci-type rules require intermediate terms to be shown because a later term may depend on more than one earlier value.
  • The examiner expects the terms in order and usually gives method credit for correct substitutions or repeated operations even if a later arithmetic error occurs.

Worked example

A sequence starts 2,52,5. Each later term is one more than the sum of the previous two terms. Find the next three terms.

  1. 1.Third term: 2+5+1=82+5+1=8.
  2. 2.Fourth term: 5+8+1=145+8+1=14.
  3. 3.Fifth term: 8+14+1=238+14+1=23.

Answer: 8,14,238,14,23.

Common mistakes

  • Don't substitute n=0n=0 for the first term when the sequence starts at n=1n=1.
  • Don't repeatedly use the original starting pair instead of the latest two terms for a two-term recurrence.

Exam tip. Write each intermediate term because a correct recurrence method can still earn marks after one arithmetic slip.

A24 · Recognise and use triangular, square and cube numbers, arithmetic progressions, Fibonacci type sequences, quadratic sequences, simple geometric progressions (r^n, r rational > 0 or a surd) and others

  • Square and cube numbers have forms n2n^2 and n3n^3, while triangular numbers have form n(n+1)2\dfrac{n(n+1)}{2}. An arithmetic progression has a constant first difference and a quadratic sequence has constant second differences.
  • Fibonacci-type sequences form later terms from preceding terms.
  • At Higher tier, a geometric progression has a constant positive rational or surd multiplier and can be written using powers such as rnr^n.
  • Check several consecutive steps before identifying a sequence, then use the defining rule consistently.
  • The examiner expects the named structure to be supported by differences, ratios or a valid term relationship rather than by visual resemblance alone.

Worked example

Higher tier: The sequence 2,23,6,63,…2,2\sqrt3,6,6\sqrt3,\ldots is geometric. Find its common ratio and eighth term.

  1. 1.Divide consecutive terms: 23÷2=32\sqrt3\div2=\sqrt3, so r=3r=\sqrt3.
  2. 2.Use un=2(3)n−1u_n=2(\sqrt3)^{n-1}.
  3. 3.u8=2(3)7=2(273)=543u_8=2(\sqrt3)^7=2(27\sqrt3)=54\sqrt3.

Answer: Common ratio 3\sqrt3; eighth term 54354\sqrt3.

Common mistakes

  • Don't call a sequence arithmetic after checking only one pair of terms.
  • Don't use first differences to identify a quadratic sequence instead of checking that second differences are constant.
  • Don't replace the exact surd ratio by a decimal and lose the exact form.

Exam tip. Show a short difference table or two equal consecutive ratios to justify the sequence type.

A25 · Deduce expressions to calculate the nth term of linear and quadratic sequences

  • For a linear sequence with common difference dd, start with dndn and adjust the constant so the expression gives the first term when n=1n=1.
  • At Higher tier, a quadratic sequence an2+bn+can^2+bn+c has constant second difference 2a2a.
  • Find aa, subtract the values of an2an^2 from the original terms, and identify the linear rule left behind to obtain bb and cc.
  • The examiner expects an expression in nn, not merely the next term.
  • Verify the rule against at least two supplied terms; this detects a shifted index or an incorrect constant before the final answer.

Worked example

Higher tier: Find the nnth term of 2,7,14,23,34,…2,7,14,23,34,\ldots.

  1. 1.First differences are 5,7,9,115,7,9,11, so the second difference is 22 and a=1a=1.
  2. 2.Subtract n2n^2 from the terms to get 1,3,5,7,91,3,5,7,9.
  3. 3.The remainder has rule 2n−12n-1, so combine the parts.

Answer: n2+2n−1n^2+2n-1.

Common mistakes

  • Don't use the constant second difference as aa instead of recognising it is 2a2a.
  • Don't use the first term itself as the constant in a linear nnth-term rule.
  • Don't find a rule that matches one term and fail to verify it against later terms.

Exam tip. For a quadratic sequence, display the first and second differences because they secure the method for finding the n2n^2 coefficient.