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Question 50

Show that the equation $x^3 + 10x = 3$ has a solution between $x = 0$ and $x = 1$ .

[2]

Show that the equation $x^3 + 10x = 3$ can be rearranged to give: $x=310−x310\displaystyle x = \frac{3}{10} - \frac{x^3}{10}$

[1]

Starting with $x_0 = 0$ , use the iteration formula $xn+1=310−xn310\displaystyle x_{n+1} = \frac{3}{10} - \frac{x_n^3}{10}$ twice to find an estimate for the solution to $x^3 + 10x = 3$

[3]

Markscheme

Iteration Questions

GCSE
Maths
Iteration

124 exam-style questions on Eduqas GCSE Maths Iteration. Each one has a worked solution and a mark scheme showing where the marks go.

Iteration

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Iteration

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Question 50

Show that the equation $x^3 + 10x = 3$ has a solution between $x = 0$ and $x = 1$ .

[2]

Show that the equation $x^3 + 10x = 3$ can be rearranged to give: $x=310−x310\displaystyle x = \frac{3}{10} - \frac{x^3}{10}$

[1]

Starting with $x_0 = 0$ , use the iteration formula $xn+1=310−xn310\displaystyle x_{n+1} = \frac{3}{10} - \frac{x_n^3}{10}$ twice to find an estimate for the solution to $x^3 + 10x = 3$

[3]

Markscheme

Iteration Questions

GCSE
Maths
Iteration

124 exam-style questions on Eduqas GCSE Maths Iteration. Each one has a worked solution and a mark scheme showing where the marks go.