Iteration for a cubic equation

Iteration

(1MA1) · Higher

ORIGINAL

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

Board and spec code confirmed against Edexcel 1MA1 · registry checked 2026-07-09How this checking works

Question

The equation

x^3+x-5=0

is rearranged as

x_{n+1}=\sqrt[3]{5-x_n}

. Starting with

x_0=1.5

, calculate

x_1

x_5

and hence estimate the root to 3 decimal places.

Every step worked, with the reasoning.

1
$x_1=\sqrt[3]{5-1.5}=1.518294\ldots$
Substitute $x_0=1.5$ into the iteration formula.
2
$x_2=\sqrt[3]{5-1.518294\ldots}=1.515644\ldots$
Feed the unrounded value of $x_1$ back into the formula.
3
$x_3=1.516029\ldots$ and $x_4=1.515973\ldots$
Continue iterating, keeping the calculator's full stored values.
4
$x_5=1.515981\ldots$
One more iteration confirms the values are settling.
5
$x_4$ and $x_5$ both round to $1.516$ (3 d.p.).
Use the stable decimal places for the requested estimate.

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

This is how to revise a method, not just read it

Fade the steps out until you can do it cold. Want a set built around exactly what you keep slipping on? Free intro call, then a free first lesson.