Pearson Edexcel · GCSE · Mathematics
Course version: 1MA1 · Version from Mar 2026
Algebra - Solving equations and inequalities: find approximate solutions to equations numerically using iteration (Higher tier only)
This is Higher tier only.
- 3 key terms
Study notes
From the checked teaching notes for this specification point.
The idea
This is Higher tier only. Iteration closes in on a root by improving a guess over and over. The equation is first rearranged so that a single x sits on the left: x² + x − 5 = 0 becomes x = √(5 − x), which gives the iterative formula xₙ₊₁ = √(5 − xₙ). Then you take the starting value the question supplies and feed each answer back in. From x₁ = 2 you get x₂ = 1.7321, x₃ = 1.8077 and x₄ = 1.7867, and the terms settle towards the root near 1.791. On a calculator, work out the first term, then use the answer key inside the formula so that one press produces the next term. Questions normally give you the rearrangement and the starting value, then ask for a stated number of terms or for the root to a given accuracy.
What earns the marks
- A rearrangement that leads to the given iterative formula when you are asked to derive it.
- A correct first iteration, with the starting value seen substituted into the formula.
- The later iterations, each carried to enough decimal places to support the answer asked for.
- The final root rounded exactly as the question states.
- Substituting two values and stating that the results have opposite signs, when asked to show a root lies between them.
Common mistakes and how to put them right
Rounding each term before feeding it back in, so the sequence settles on a value that is not the root.
Typing the starting value in again for every term instead of using the previous answer.
Labelling the terms from the wrong place, so the value called x₃ is really the second iteration.
Giving the root to a different accuracy from the one the question asked for.
Exam tips
- Keep the full calculator display for every term and round only at the very end.
- Write each term on its own line with its label, so the examiner can see which iteration is which.
- If you are asked to show that a root lies between two numbers, substitute both into the equation and say that the two answers have opposite signs.
Try the lesson
The start of the Algebra - Solving equations and inequalities: find approximate solutions to equations numerically using iteration (Higher tier only) lesson, made from checked teaching notes for this specification point.
Have a go first
The equation x³ + 4x - 8 = 0 has a solution near x = 1.5. Using the iteration formula xₙ₊₁ = (8 − 4xₙ)^(1/3) with x₀ = 1.5, work out the value of x₂. Give your answer correct to 3 decimal places.
Listen
This is Higher tier only. Iteration closes in on a root by improving a guess over and over. The equation is first rearranged so that a single x sits on the left: x² + x − 5 = 0 becomes x = √(5 − x), which gives the iterative formula xₙ₊₁ = √(5 − xₙ). Then you take the starting value the question supplies and feed each answer back in. From x₁ = 2 you get x₂ = 1.7321, x₃ = 1.8077 and x₄ = 1.7867, and the terms settle towards the root near 1.791. On a calculator, work out the first term, then use the answer key inside the formula so that one press produces the next term. Questions normally give you the rearrangement and the starting value, then ask for a stated number of terms or for the root to a given accuracy. A strong answer earns marks for these points. a rearrangement that leads to the given iterative formula when you are asked to derive it. a correct first iteration, with the starting value seen substituted into the formula. the later iterations, each carried to enough decimal places to support the answer asked for. the final root rounded exactly as the question states. Exam tip: Keep the full calculator display for every term and round only at the very end.
Key words
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iterative methods
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iteration formula
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successive approximations
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See every definition
- iterative methods
- Ways of finding approximate solutions to equations numerically, closing in on a root by improving a guess over and over.
- iteration formula
- A rearrangement with a single x on the left, such as xₙ₊₁ = √(5 − xₙ), into which each answer is fed back.
- successive approximations
- The terms produced by taking the starting value the question supplies and feeding each answer back in, settling towards the root.
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