Mathematics 9709 · syllabus 2.6, 3.6
Cambridge A Level Mathematics 9709 — Numerical solution of equations topical past papers
Locating a root by a change of sign, then finding it to a stated accuracy with an iterative formula xₙ₊₁ = F(xₙ) when algebra cannot solve the equation.
- Questions
- 105
- Years
- 2009–2026
- With mark scheme
- 100%
- Questions 2022–2026
- 30
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What Numerical solution of equations covers
Some equations, such as x = 2 cos x or eˣ = 4 − x, cannot be solved exactly. Papers 2 and 3 (sections 2.6 and 3.6, which have the same content) ask you to locate a root approximately, by sketching two graphs to see where they cross or by showing that a continuous function changes sign between two values, and then to pin the root down with an iteration.
An iterative formula xₙ₊₁ = F(xₙ) comes from rearranging the equation into the form x = F(x). Starting from a value near the root, you apply the formula repeatedly until successive values agree to the accuracy required. You need to show how a given formula relates to the equation, and to know that an iteration can fail to converge; the condition for convergence itself is not examined.
Questions are usually in several parts and draw on other topics: the equation often comes from a stationary point, an area or a sector, so differentiation, integration or circular measure produces the equation before the numerical method solves it. A typical sequence is to form the equation, show that a root lies in an interval, confirm that an iterative formula comes from the equation, and then find the root. The equations can involve any function in Papers 2 and 3, including ln x, eˣ and the trigonometric functions.
In the syllabus
The headings this topic has in the Cambridge International 2026–2027 syllabus, as the board prints them.
- 2.6 Numerical solution of equations
- 3.6 Numerical solution of equations
What you should be able to do
- Show by a change of sign that a root lies in a given interval
- Sketch two graphs to show how many roots an equation has
- Show that an equation can be rearranged into a given iterative form
- Use an iterative formula to find a root to a stated accuracy
- Confirm a root to a given number of decimal places with a sign change over the rounding interval
Key ideas
- If f is continuous on an interval and f(a) and f(b) have opposite signs, then f(x) = 0 has a root between a and b
- A root of x = F(x) is where the line y = x meets the curve y = F(x); if xₙ₊₁ = F(xₙ) converges, it converges to such a root
- An iteration can diverge, or settle on a different root, depending on the rearrangement and the starting value
- A root is α to 2 decimal places if the function changes sign over [α − 0.005, α + 0.005]
How Numerical solution of equations has been examined
From 2022 to 2026, Numerical solution of equations was on Paper 3 · Pure Mathematics 3 in 11 of 11 exam series, worth about 10% of that paper’s marks.
| Paper | Questions | Share of marks | Series |
|---|---|---|---|
| Paper 3 · Pure Mathematics 3 | 30 | 10% | 11 of 11 |
Momentum’s own count, over the questions in its bank for these years: a series is one exam session, every zone included, and a question filed under two chapters counts in both. Median marks per question: 8.
Filed in Momentum under: Numerical Methods.
A worked example
Show that x³ + 2x − 7 = 0 has a root between 1 and 2. Use the iteration xₙ₊₁ = ∛(7 − 2xₙ), starting with x₁ = 1.5, to find the root correct to 2 decimal places.
- Let f(x) = x³ + 2x − 7. f(1) = −4 and f(2) = 5; f is continuous and changes sign, so a root lies between 1 and 2.
- The formula comes from the equation: x³ = 7 − 2x, so x = ∛(7 − 2x).
- Iterate from x₁ = 1.5: x₂ = 1.5874, x₃ = 1.5639, x₄ = 1.5703, x₅ = 1.5686, x₆ = 1.5690, x₇ = 1.5689.
- The iterates now agree to 2 decimal places, at 1.57.
- Check: f(1.565) = −0.037 and f(1.575) = 0.057 have opposite signs, so the root is 1.57 to 2 decimal places.
Answer: 1.57 (2 d.p.)
Written by Momentum for this page, with its own numbers; it is not taken from a past paper.
Mistakes that cost marks
- Writing only 'there is a sign change': the two function values and the conclusion 'so a root lies between a and b' are what earn the marks.
- Relying on a change of sign across a discontinuity, as with tan x at π/2: the function must be continuous on the interval for the argument to work.
- Rounding each iterate before feeding it back: carry full calculator values and round only the final answer.
- Writing iterates only to the accuracy of the final answer: show each to at least one more decimal place than required, so it is clear the sequence has settled.
- Iterating a formula that contains sin x or cos x in degree mode: such iterations are almost always in radians.
Exam technique
- When asked to show that a root satisfies an equation, rearrange step by step from the given equation to x = F(x), or the reverse, without skipping lines.
- Use the calculator's answer key to repeat the formula, and write down each iterate to 4 or more decimal places.
- Give the final answer to the accuracy asked and, where needed, justify it with a sign change over the rounding interval.
Practising in Momentum
- Every question in a chapter as one list, from every paper and year in the bank, newest first or shuffled.
- Each question turns over to its own part of the mark scheme, where the board published one.
- Narrow by paper, year, session, zone and marks, and star the questions worth keeping.
- Pick chapters, a number of questions and a time, and the exam builder draws a mock to sit and mark.
Past-paper questions and mark schemes are the copyright of Cambridge International. Momentum shows them to signed-in learners for revision and does not publish them on these pages.