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AlgebraHigher

Iteration GCSE questions with answers

6 iteration questions to try, each with the answer and the full working one tap away. Every question here was made and worked out by the Show Working maths engine, not typed in by hand.

Iteration questions on Higher give you a formula like xₙ₊₁ = ∛(14 + 2xₙ) and a starting value. You put each answer back into the formula to get the next one, which gets closer and closer to a solution of an equation.

Quiz yourself: 6 questions

Worked example 1: iteration

Use the iteration formula xₙ₊₁ = √3 + 2xₙ with x₀ = 1.
Find the values of x₁, x₂ and x₃. Give each one correct to 3 decimal places.

You can use a calculator for this one.

  1. Put x₀ = 1 into the formula: x₁ = √3 + 2 × 1 = 2.236067..., which is 2.236 to 3 decimal places.
  2. Put that answer back in (use the full value, with the ANS key): x₂ = √3 + 2 × x₁ = 2.733520..., so x₂ = 2.734.
  3. Once more: x₃ = √3 + 2 × x₂ = 2.909818..., so x₃ = 2.910.

Answer x₁ = 2.236, x₂ = 2.734, x₃ = 2.910

01 / Quiz yourselfTry it, then tap to check

Iteration questions to try

Work each one out on paper first. Then open the answer and compare your working with ours, step by step.

01 / IterationHigher only

Use the iteration formula xₙ₊₁ = √8 + 2xₙ with x₀ = 3.
Find the values of x₁, x₂ and x₃. Give each one correct to 3 decimal places.

You can use a calculator for this one.

Show the answer and working
  1. Put x₀ = 3 into the formula: x₁ = √8 + 2 × 3 = 3.741657..., which is 3.742 to 3 decimal places.
  2. Put that answer back in (use the full value, with the ANS key): x₂ = √8 + 2 × x₁ = 3.934884..., so x₂ = 3.935.
  3. Once more: x₃ = √8 + 2 × x₂ = 3.983687..., so x₃ = 3.984.

Answer x₁ = 3.742, x₂ = 3.935, x₃ = 3.984

02 / IterationHigher only

Use the iteration formula xₙ₊₁ = 2xₙ2 + 4 with x₀ = 0.
Find the values of x₁, x₂ and x₃. Give each one correct to 3 decimal places.

You can use a calculator for this one.

Show the answer and working
  1. Put x₀ = 0 into the formula: x₁ = 202 + 4 = 0.500000..., which is 0.500 to 3 decimal places.
  2. Put that answer back in (use the full value, with the ANS key): x₂ = 2x₁2 + 4 = 0.470588..., so x₂ = 0.471.
  3. Once more: x₃ = 2x₂2 + 4 = 0.473770..., so x₃ = 0.474.

Answer x₁ = 0.500, x₂ = 0.471, x₃ = 0.474

03 / IterationHigher only

Use the iteration formula xₙ₊₁ = ∛(20 − xₙ) with x₀ = 2.
Find the values of x₁, x₂ and x₃. Give each one correct to 3 decimal places.

You can use a calculator for this one.

Show the answer and working
  1. Put x₀ = 2 into the formula: x₁ = ∛(20 − 1 × 2) = 2.620741..., which is 2.621 to 3 decimal places.
  2. Put that answer back in (use the full value, with the ANS key): x₂ = ∛(20 − 1 × x₁) = 2.590262..., so x₂ = 2.590.
  3. Once more: x₃ = ∛(20 − 1 × x₂) = 2.591775..., so x₃ = 2.592.

Answer x₁ = 2.621, x₂ = 2.590, x₃ = 2.592

04 / IterationHigher only

Use the iteration formula xₙ₊₁ = √13 + 4xₙ with x₀ = 1.
Find the values of x₁, x₂ and x₃. Give each one correct to 3 decimal places.

You can use a calculator for this one.

Show the answer and working
  1. Put x₀ = 1 into the formula: x₁ = √13 + 4 × 1 = 4.123105..., which is 4.123 to 3 decimal places.
  2. Put that answer back in (use the full value, with the ANS key): x₂ = √13 + 4 × x₁ = 5.430692..., so x₂ = 5.431.
  3. Once more: x₃ = √13 + 4 × x₂ = 5.892603..., so x₃ = 5.893.

Answer x₁ = 4.123, x₂ = 5.431, x₃ = 5.893

05 / IterationHigher only

Use the iteration formula xₙ₊₁ = √4 + 4xₙ with x₀ = 2.
Find the values of x₁, x₂ and x₃. Give each one correct to 3 decimal places.

You can use a calculator for this one.

Show the answer and working
  1. Put x₀ = 2 into the formula: x₁ = √4 + 4 × 2 = 3.464101..., which is 3.464 to 3 decimal places.
  2. Put that answer back in (use the full value, with the ANS key): x₂ = √4 + 4 × x₁ = 4.225684..., so x₂ = 4.226.
  3. Once more: x₃ = √4 + 4 × x₂ = 4.571951..., so x₃ = 4.572.

Answer x₁ = 3.464, x₂ = 4.226, x₃ = 4.572

06 / IterationHigher only

Use the iteration formula xₙ₊₁ = ∛(6 − 3xₙ) with x₀ = 0.
Find the values of x₁, x₂ and x₃. Give each one correct to 3 decimal places.

You can use a calculator for this one.

Show the answer and working
  1. Put x₀ = 0 into the formula: x₁ = ∛(6 − 3 × 0) = 1.817120..., which is 1.817 to 3 decimal places.
  2. Put that answer back in (use the full value, with the ANS key): x₂ = ∛(6 − 3 × x₁) = 0.818644..., so x₂ = 0.819.
  3. Once more: x₃ = ∛(6 − 3 × x₂) = 1.524639..., so x₃ = 1.525.

Answer x₁ = 1.817, x₂ = 0.819, x₃ = 1.525

02 / What's covered1 kind of question

Iteration at GCSE and where marks go

  • IterationHigher only

Worked example 2: iteration

Use the iteration formula xₙ₊₁ = 3xₙ2 + 3 with x₀ = 2.
Find the values of x₁, x₂ and x₃. Give each one correct to 3 decimal places.

You can use a calculator for this one.

  1. Put x₀ = 2 into the formula: x₁ = 322 + 3 = 0.428571..., which is 0.429 to 3 decimal places.
  2. Put that answer back in (use the full value, with the ANS key): x₂ = 3x₁2 + 3 = 0.942307..., so x₂ = 0.942.
  3. Once more: x₃ = 3x₂2 + 3 = 0.771616..., so x₃ = 0.772.

Answer x₁ = 0.429, x₂ = 0.942, x₃ = 0.772

A common mistake

Rounding each value before putting it back in. Keep the full calculator value for the next step and only round what you write down. The ANS key does this for you.

03 / Practise itiPhone and iPad

More iteration questions in Show Working

In the app, Iteration makes fresh questions every time, so they never run out. Tap Show me one to see a new worked example straight away, or practise ten and open the working whenever you are stuck. Forgotten a method? Type it into Look up and the answer is there.

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