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GCSE Maths · Edexcel 1MA1 · Higher · Grades 7–9 · Algebra

Iteration and Numerical Methods

Some equations, like x³ − 3x − 5 = 0, can’t be solved by factorising or the formula. Iteration finds the answer to any accuracy by feeding each answer back into a formula. It is a Higher-only topic, usually a 5–6 mark question with three parts that always follow the same pattern.

1Step 1: show there is a root between two values

If f(x) changes sign between x = a and x = b (and the graph is continuous), the graph must cross zero, so there is a root between a and b.

Worked example
Show that x³ − 3x − 5 = 0 has a solution between x = 2 and x = 3.
  1. f(2) = 8 − 6 − 5 = −3 (negative).
  2. f(3) = 27 − 9 − 5 = 13 (positive).
  3. There is a change of sign, so there is a root between 2 and 3.
Sign change ⇒ root between 2 and 3
Exam tipYou must write both values and the conclusion “there is a change of sign, so there is a root”.

2Step 2: rearrange into x = …

Worked example
Show that x³ − 3x − 5 = 0 can be rearranged to give x = ∛(3x + 5).
  1. x³ − 3x − 5 = 0
  2. x³ = 3x + 5
  3. x = ∛(3x + 5)
Shown

3Step 3: iterate

Turn the rearrangement into an iterative formula, xn+1 = ∛(3xn + 5), and feed each answer back in.

Worked example
Starting with x0 = 2, find x1, x2 and x3 to 4 decimal places.
  1. x1 = ∛(3 × 2 + 5) = ∛11 = 2.2240
  2. x2 = ∛(3 × 2.2240 + 5) = 2.2684
  3. x3 = ∛(3 × 2.2684 + 5) = 2.2770
x₁ = 2.2240, x₂ = 2.2684, x₃ = 2.2770
Calculator tipType 2 and press =. Then type ∛(3 × Ans + 5) and press = repeatedly. Each press gives the next term. Write down each value to 4 d.p., but keep the full value in Ans.

4What the answer means

The values are getting closer and closer to the root, about 2.279. To show a root is correct to 2 d.p., check the sign of f at the bounds:

Worked example
Show the root is 2.28 to 2 decimal places.
  1. f(2.275) = -0.0505 (negative).
  2. f(2.285) = 0.0755 (positive).
  3. There is a change of sign, so the root lies between 2.275 and 2.285, and rounds to 2.28.
Root = 2.28 (2 d.p.)
Exam tip“What does x₃ represent?” Answer: an estimate of the solution of the equation x³ − 3x − 5 = 0.

5Iterative processes in context

The same idea models repeated growth or decay. For example, Pn+1 = 1.05Pn is 5% compound interest each year.

Worked example
£2000 is invested with Pn+1 = 1.05Pn and P0 = 2000. Find P3.
  1. P₁ = 2100.00
  2. P₂ = 2205.00
  3. P₃ = 2315.25
£2315.25
Common mistakes
  • Showing f(a) and f(b) without stating that the sign changes, so there is a root.
  • Rounding each iterate before feeding it back in. Use the Ans key.
  • Using degrees instead of numbers by mistake, or mistyping the cube root.
  • Showing a rearrangement by working backwards from the answer. Start from the original equation.

★Key Vocabulary Recap

Iteration
Repeating a process, using each output as the next input.
Iterative formula
A rule like xn+1 = g(xn).
Converge
Get closer and closer to a fixed value.
Root
A solution of f(x) = 0.
Change of sign
f(a) and f(b) have opposite signs.
Specification reference (Pearson Edexcel GCSE (9–1) Mathematics 1MA1: A20, R16 (iterative processes))
  • A20: find approximate solutions to equations numerically using iteration.
  • R16: set up, solve and interpret the answers in growth and decay problems, including compound interest, and work with general iterative processes.
Next stepReady to practise? Try the 9 exam-style questions, then check your working against the worked solutions.
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Practice questions

GCSE Maths · Edexcel 1MA1 · Higher · Grades 7–9 · Algebra

Iteration and Numerical Methods — Practice Questions

9 exam-style questions (22 marks), building from warm-up to grade 8–9. Show full working: Edexcel gives method marks even when the final answer is wrong.

Higher · Grades 7–9
  1. (2 marks)
    Grade 7
    Show that x³ + 2x − 7 = 0 has a solution between x = 1 and x = 2.
  2. (2 marks)
    Grade 7
    Show that x³ + 2x − 7 = 0 can be rearranged to give x = ∛(7 − 2x).
  3. (3 marks)
    Grade 7
    Using xn+1 = ∛(7 − 2xn) with x0 = 1.5, find x1, x2 and x3, giving each to 4 decimal places.
  4. (2 marks)
    Grade 7
    Using xn+1 = 2 + 3xn with x0 = 1, find x1, x2 and x3.
  5. (2 marks)
    Grade 8
    The iteration in Question 4 converges. Show that its limit L satisfies L² − 2L − 3 = 0, and find L.
  6. (3 marks)
    Grade 8
    Use xn+1 = xn² + 25 with x0 = 0 to find x3 to 4 d.p. Which equation is this iteration solving?
  7. (3 marks)
    Grade 8
    (a) Show that x³ − 6x + 2 = 0 can be written as x = √(6 − 2x).
    (b) Use x0 = 2 to find x3 to 4 d.p.
  8. (3 marks)
    Grade 9
    Show that the equation x³ − 3x − 5 = 0 has a root that is 2.28 correct to 2 decimal places.
  9. (2 marks)
    Grade 7
    A population of bacteria follows Pn+1 = 1.05Pn, where n is the number of hours. P0 = 2000. Find P3 to the nearest whole number.
When you’re doneRevise the method in the notes, then mark your answers with the worked solutions.
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Worked solutions

Show worked solutions
GCSE Maths · Edexcel 1MA1 · Higher · Grades 7–9 · Algebra

Iteration and Numerical Methods — Worked Solutions

Full worked solutions to all 9 questions (22 marks), set out the way an Edexcel examiner wants to see them.

Higher · Grades 7–9
  1. (2 marks)
    Grade 7
    Show that x³ + 2x − 7 = 0 has a solution between x = 1 and x = 2.
    Solution
    1. f(1) = 1 + 2 − 7 = −4.
    2. f(2) = 8 + 4 − 7 = 5.
    3. Change of sign, so there is a root between 1 and 2.
    Sign change from −4 to 5
  2. (2 marks)
    Grade 7
    Show that x³ + 2x − 7 = 0 can be rearranged to give x = ∛(7 − 2x).
    Solution
    1. x³ = 7 − 2x
    2. x = ∛(7 − 2x)
    Shown
  3. (3 marks)
    Grade 7
    Using xn+1 = ∛(7 − 2xn) with x0 = 1.5, find x1, x2 and x3, giving each to 4 decimal places.
    Solution
    1. x₁ = ∛(7 − 3) = ∛4 = 1.5874
    2. x₂ = ∛(7 − 2 × 1.5874) = 1.5639
    3. x₃ = 1.5703
    1.5874, 1.5639, 1.5703
  4. (2 marks)
    Grade 7
    Using xn+1 = 2 + 3xn with x0 = 1, find x1, x2 and x3.
    Solution
    1. x₁ = 2 + 3 = 5.
    2. x₂ = 2 + 3/5 = 2.6.
    3. x₃ = 2 + 3/2.6 = 3.1538
    5, 2.6, 3.1538
  5. (2 marks)
    Grade 8
    The iteration in Question 4 converges. Show that its limit L satisfies L² − 2L − 3 = 0, and find L.
    Solution
    1. At the limit, L = 2 + 3/L.
    2. Multiply by L: L² = 2L + 3, so L² − 2L − 3 = 0.
    3. (L − 3)(L + 1) = 0. The terms are positive, so L = 3.
    L = 3
  6. (3 marks)
    Grade 8
    Use xn+1 = xn² + 25 with x0 = 0 to find x3 to 4 d.p. Which equation is this iteration solving?
    Solution
    1. x₁ = 2/5 = 0.4; x₂ = (0.16 + 2)/5 = 0.432; x₃ = 0.4373.
    2. x = (x² + 2)/5 → 5x = x² + 2 → x² − 5x + 2 = 0.
    x₃ = 0.4373; x² − 5x + 2 = 0 (root ≈ 0.4384)
  7. (3 marks)
    Grade 8
    (a) Show that x³ − 6x + 2 = 0 can be written as x = √(6 − 2x).
    (b) Use x0 = 2 to find x3 to 4 d.p.
    Solution
    1. (a) x³ = 6x − 2. Divide by x: x² = 6 − 2/x, so x = √(6 − 2/x).
    2. (b) x₁ = √5 = 2.2361; x₂ = 2.2596; x₃ = 2.2616.
    x₃ = 2.2616
  8. (3 marks)
    Grade 9
    Show that the equation x³ − 3x − 5 = 0 has a root that is 2.28 correct to 2 decimal places.
    Solution
    1. f(2.275) = -0.0505 < 0
    2. f(2.285) = 0.0755 > 0
    3. Change of sign between the bounds 2.275 and 2.285, so the root is 2.28 to 2 d.p.
    Sign change between 2.275 and 2.285
  9. (2 marks)
    Grade 7
    A population of bacteria follows Pn+1 = 1.05Pn, where n is the number of hours. P0 = 2000. Find P3 to the nearest whole number.
    Solution
    1. P₁ = 2100, P₂ = 2205, P₃ = 2315.25.
    2315
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