GCSE · Iteration

Showing a root exists — change of signExam-style questions

Why a sign change in a continuous function traps a root between two values, how to present the argument for full marks, and how the same idea justifies a root to a given accuracy.

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13 questions · GCSE (Higher) level · mark scheme and worked solution on each · no sign-up to try

DifficultyAll Level 3 Level 4 Level 5

1
(a) Show that the equation $x^3 + 2x = 15$ has a solution between $x = 2$ and $x = 3$ (2 marks)

(b) Starting with $x_0 = 2$ , use the iteration $x_{n+1} = \sqrt[3]{15 - 2x_n}$ to find $x_3$ . Give your answer correct to 3 decimal places. (2 marks)
Level 3 · 4 marksTry it
2
(a) Show that the equation $x^3 - 5x + 1 = 0$ has a solution between $x = 2$ and $x = 3$ (2 marks)
(b) The equation $x^3 - 5x + 1 = 0$ can be rearranged to give $x = \sqrt[3]{5x - 1}$

Starting with $x_0 = 2$ , use the iteration formula $x_{n+1} = \sqrt[3]{5x_n - 1}$ three times to find an estimate for the solution.

Give your answer correct to 3 decimal places. (2 marks)
Level 3 · 4 marksTry it
3
(a) Show that the equation $x^3 + x - 7 = 0$ has a solution between $x = 1$ and $x = 2$ . [2]

(b) Show that the equation can be rearranged to $x = \sqrt[3]{7 - x}$ . [1]

(c) Starting with $x_0 = 1.5$ , use the iteration $x_{n+1} = \sqrt[3]{7 - x_n}$ to find $x_3$ . Give your answer correct to 3 decimal places. [2]
Level 3 · 5 marksTry it
4
$f(x) = x^3 + x - 20$

Show that the equation $f(x) = 0$ has a solution between $x = 2$ and $x = 3$
Level 3 · 2 marksTry it
5
Show that the equation $x^3 - 7x - 8 = 0$ has a solution between $x = 3$ and $x = 4$
Level 3 · 2 marksTry it

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