GCSE & IGCSE maths practice

Iteration questions and answers

Iteration finds an approximate solution to an equation you can't solve exactly. You show a solution lies between two values (a change of sign), rearrange the equation into the form $x = g(x)$ , then use $x_{n+1} = g(x_n)$ again and again. Each answer feeds into the next. It's GCSE Higher only. Use the ANS key on your calculator so you don't lose accuracy.

9 questions · Higher tier · Every answer is checked as you type it, with a full worked solution. Updated 26 Oct 2026.

Start the questions nth term practice (subscript notation)

Where iteration comes up

GCSE (AQA, Edexcel, OCR)	Higher only: finding approximate solutions to equations numerically using iteration.
Cambridge IGCSE (0580)	Not in the 2025 to 2027 syllabus.
Edexcel International GCSE (4MA1)	Not in the specification.

The method

To show a root lies between a and b: substitute both into f(x) (with everything on one side) and show the signs are different.
To show a rearrangement: start from the equation and rearrange step by step until you reach x = g(x).
Type the starting value into your calculator, press =, then type the formula using ANS in place of xₙ. Each = gives the next value.
Write down each value to more decimal places than the question asks for, and only round at the end.

Iteration questions

Try each one before looking at the hint. If you're stuck, the AI tutor can talk you through it step by step.

Warm-up

Question 1Warm-upHigher

Using $x_{n+1} = 2 + \dfrac{1}{x_n}$ with $x_0 = 2$ , find $x_1$ , $x_2$ and $x_3$ . Give $x_3$ to 4 decimal places.

Question 2Warm-upHigher

Using $x_{n+1} = \sqrt{5 + x_n}$ with $x_0 = 2$ , find $x_1$ , $x_2$ and $x_3$ to 4 decimal places.

Exam-style

Question 3Exam-styleHigher

Show that the equation $x^3 + x - 4 = 0$ has a solution between $x = 1$ and $x = 2$ .

Question 4Exam-styleHigher

Show that $x^3 + x - 4 = 0$ can be rearranged to give $x = \sqrt[3]{4 - x}$ .

Question 5Exam-styleHigher

Using $x_{n+1} = \sqrt[3]{4 - x_n}$ with $x_0 = 1.5$ , find $x_1$ , $x_2$ and $x_3$ to 4 decimal places.

Question 6Exam-styleHigher

Show that $x^3 - 3x - 5 = 0$ has a solution between $x = 2$ and $x = 3$ .

Question 7Exam-styleHigher

Using $x_{n+1} = \sqrt[3]{3x_n + 5}$ with $x_0 = 2$ , find $x_1$ , $x_2$ and $x_3$ to 4 decimal places.

Challenge

Question 8ChallengeHigher

The equation $x^2 - 5x + 3 = 0$ can be written as $x = \dfrac{x^2 + 3}{5}$ . Using $x_{n+1} = \dfrac{x_n^2 + 3}{5}$ with $x_0 = 1$ , find $x_1$ , $x_2$ and $x_3$ . Give $x_3$ to 4 decimal places.

Question 9ChallengeHigher

Keep iterating $x_{n+1} = \sqrt[3]{3x_n + 5}$ from $x_0 = 2$ until the values agree to 2 decimal places. What is the solution of $x^3 - 3x - 5 = 0$ to 2 decimal places?

Common mistakes

Rounding each iteration before using it, so later values drift.
Stopping at the wrong subscript: x₃ is the third value after x₀.
Showing a sign change with only one value substituted, or not saying 'so there's a root between them'.

Stuck on one of these?

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Sources

The questions are our own. The “where it comes up” notes are from each board's published specification, checked on 26 October 2026:

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