All worked examples

Iteration for a cubic equation

Iteration
(1MA1) · Higher
ORIGINAL

Apply a recurrence accurately and use stabilising decimal places to estimate a cubic root.

Verified against Edexcel 1MA1 (2026 spec)

Question

The equation x3+x−5=0x^3+x-5=0 is rearranged as xn+1=5−xn3x_{n+1}=\sqrt[3]{5-x_n}. Starting with x0=1.5x_0=1.5, calculate x1x_1 to x5x_5 and hence estimate the root to 3 decimal places.

Every step worked, with the reasoning.

  1. 1
    x1=5−1.53=1.518294…x_1=\sqrt[3]{5-1.5}=1.518294\ldots

    Substitute x0=1.5x_0=1.5 into the iteration formula.

  2. 2
    x2=5−1.518294…3=1.515644…x_2=\sqrt[3]{5-1.518294\ldots}=1.515644\ldots

    Feed the unrounded value of x1x_1 back into the formula.

  3. 3
    x3=1.516029…x_3=1.516029\ldots and x4=1.515973…x_4=1.515973\ldots

    Continue iterating, keeping the calculator's full stored values.

  4. 4
    x5=1.515981…x_5=1.515981\ldots

    One more iteration confirms the values are settling.

  5. 5
    x4x_4 and x5x_5 both round to 1.5161.516 (3 d.p.).

    Use the stable decimal places for the requested estimate.

Answer: x≈1.516x \approx 1.516 (3 d.p.)

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