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Iteration
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Level 3
Level 4
Level 5
Showing a root exists — change of sign
13 questions
Lesson
Not started
Mark as done
f
(
x
)
=
x
3
+
x
−
20
f(x) = x^3 + x - 20
f
(
x
)
=
x
3
+
x
−
20
Show that the equation
f
(
x
)
=
0
f(x) = 0
f
(
x
)
=
0
has a solution between
x
=
2
x = 2
x
=
2
and
x
=
3
x = 3
x
=
3
●●●
●●
Level 3
2 marks
Start
→
Mark as done
Show that the equation
x
3
−
7
x
−
8
=
0
x^3 - 7x - 8 = 0
x
3
−
7
x
−
8
=
0
has a solution between
x
=
3
x = 3
x
=
3
and
x
=
4
x = 4
x
=
4
●●●
●●
Level 3
2 marks
Start
→
Mark as done
f
(
x
)
=
x
3
−
4
x
−
9
f(x) = x^3 - 4x - 9
f
(
x
)
=
x
3
−
4
x
−
9
Show that the equation
f
(
x
)
=
0
f(x) = 0
f
(
x
)
=
0
has a solution between
x
=
2
x = 2
x
=
2
and
x
=
3
x = 3
x
=
3
●●●
●●
Level 3
3 marks
Start
→
Mark as done
(a)
Show that the equation
x
3
+
2
x
=
15
x^3 + 2x = 15
x
3
+
2
x
=
15
has a solution between
x
=
2
x = 2
x
=
2
and
x
=
3
x = 3
x
=
3
(2 marks)
(b)
Starting with
x
0
=
2
x_0 = 2
x
0
=
2
, use the iteration
x
n
+
1
=
15
−
2
x
n
3
x_{n+1} = \sqrt[3]{15 - 2x_n}
x
n
+
1
=
3
15
−
2
x
n
to find
x
3
x_3
x
3
. Give your answer correct to 3 decimal places. (2 marks)
●●●
●●
Level 3
4 marks
Start
→
Mark as done
(a)
Show that the equation
x
3
−
5
x
+
1
=
0
x^3 - 5x + 1 = 0
x
3
−
5
x
+
1
=
0
has a solution between
x
=
2
x = 2
x
=
2
and
x
=
3
x = 3
x
=
3
(2 marks)
(b)
The equation
x
3
−
5
x
+
1
=
0
x^3 - 5x + 1 = 0
x
3
−
5
x
+
1
=
0
can be rearranged to give
x
=
5
x
−
1
3
x = \sqrt[3]{5x - 1}
x
=
3
5
x
−
1
Starting with
x
0
=
2
x_0 = 2
x
0
=
2
, use the iteration formula
x
n
+
1
=
5
x
n
−
1
3
x_{n+1} = \sqrt[3]{5x_n - 1}
x
n
+
1
=
3
5
x
n
−
1
three times to find an estimate for the solution.
Give your answer correct to 3 decimal places. (2 marks)
●●●
●●
Level 3
4 marks
Start
→
Mark as done
(a)
Show that the equation
x
3
+
x
−
7
=
0
x^3 + x - 7 = 0
x
3
+
x
−
7
=
0
has a solution between
x
=
1
x = 1
x
=
1
and
x
=
2
x = 2
x
=
2
. [2]
(b)
Show that the equation can be rearranged to
x
=
7
−
x
3
x = \sqrt[3]{7 - x}
x
=
3
7
−
x
. [1]
(c)
Starting with
x
0
=
1.5
x_0 = 1.5
x
0
=
1.5
, use the iteration
x
n
+
1
=
7
−
x
n
3
x_{n+1} = \sqrt[3]{7 - x_n}
x
n
+
1
=
3
7
−
x
n
to find
x
3
x_3
x
3
. Give your answer correct to 3 decimal places. [2]
●●●
●●
Level 3
5 marks
Start
→
Mark as done
Show that the equation
x
3
−
5
x
+
1
=
0
x^3 - 5x + 1 = 0
x
3
−
5
x
+
1
=
0
has a solution between
x
=
0
x = 0
x
=
0
and
x
=
1
x = 1
x
=
1
●●●●
●
Level 4
2 marks
Start
→
Mark as done
f
(
x
)
=
x
3
+
x
−
20
f(x) = x^3 + x - 20
f
(
x
)
=
x
3
+
x
−
20
Show that the equation
f
(
x
)
=
0
f(x) = 0
f
(
x
)
=
0
has a solution between
x
=
2.5
x = 2.5
x
=
2.5
and
x
=
2.6
x = 2.6
x
=
2.6
●●●●
●
Level 4
2 marks
Start
→
Mark as done
Show that the equation
x
3
+
2
x
−
30
=
0
x^3 + 2x - 30 = 0
x
3
+
2
x
−
30
=
0
has a solution between
x
=
2
x = 2
x
=
2
and
x
=
3
x = 3
x
=
3
●●●●
●
Level 4
3 marks
Start
→
Mark as done
f
(
x
)
=
x
3
−
4
x
−
9
f(x) = x^3 - 4x - 9
f
(
x
)
=
x
3
−
4
x
−
9
Show that the equation
f
(
x
)
=
0
f(x) = 0
f
(
x
)
=
0
has a solution between
x
=
2.7
x = 2.7
x
=
2.7
and
x
=
2.8
x = 2.8
x
=
2.8
●●●●
●
Level 4
3 marks
Start
→
Mark as done
f
(
x
)
=
x
2
−
6
x
+
2
f(x) = x^2 - 6x + 2
f
(
x
)
=
x
2
−
6
x
+
2
(a)
Show that the equation
f
(
x
)
=
0
f(x) = 0
f
(
x
)
=
0
has a solution between
x
=
0
x = 0
x
=
0
and
x
=
1
x = 1
x
=
1
(b)
Show that the equation has a
second
solution between
x
=
5
x = 5
x
=
5
and
x
=
6
x = 6
x
=
6
●●●●
●
Level 4
4 marks
Start
→
Mark as done
f
(
x
)
=
x
3
+
x
−
20
f(x) = x^3 + x - 20
f
(
x
)
=
x
3
+
x
−
20
The equation
f
(
x
)
=
0
f(x) = 0
f
(
x
)
=
0
has a solution close to
x
=
2.6
x = 2.6
x
=
2.6
Show that this solution is
x
=
2.59
x = 2.59
x
=
2.59
, correct to 2 decimal places.
●●●●●
Level 5
3 marks
Start
→
Mark as done
f
(
x
)
=
x
3
−
4
x
−
9
f(x) = x^3 - 4x - 9
f
(
x
)
=
x
3
−
4
x
−
9
The equation
f
(
x
)
=
0
f(x) = 0
f
(
x
)
=
0
has a solution close to
x
=
2.7
x = 2.7
x
=
2.7
Show that this solution is
2.7
2.7
2.7
correct to 1 decimal place
.
●●●●●
Level 5
3 marks
Start
→
Using an iterative formula
15 questions
Lesson
Not started
Mark as done
(a)
Show that the equation
x
3
+
2
x
=
15
x^3 + 2x = 15
x
3
+
2
x
=
15
has a solution between
x
=
2
x = 2
x
=
2
and
x
=
3
x = 3
x
=
3
(2 marks)
(b)
Starting with
x
0
=
2
x_0 = 2
x
0
=
2
, use the iteration
x
n
+
1
=
15
−
2
x
n
3
x_{n+1} = \sqrt[3]{15 - 2x_n}
x
n
+
1
=
3
15
−
2
x
n
to find
x
3
x_3
x
3
. Give your answer correct to 3 decimal places. (2 marks)
●●●
●●
Level 3
4 marks
Start
→
Mark as done
(a)
Show that the equation
x
3
−
3
x
−
5
=
0
x^3 - 3x - 5 = 0
x
3
−
3
x
−
5
=
0
can be rearranged to give
x
=
3
x
+
5
3
x = \sqrt[3]{3x + 5}
x
=
3
3
x
+
5
(1 mark)
(b)
Starting with
x
0
=
2
x_0 = 2
x
0
=
2
, use the iteration
x
n
+
1
=
3
x
n
+
5
3
x_{n+1} = \sqrt[3]{3x_n + 5}
x
n
+
1
=
3
3
x
n
+
5
to find
x
1
x_1
x
1
,
x
2
x_2
x
2
and
x
3
x_3
x
3
. Give your answers correct to 4 decimal places. (2 marks)
(c)
Write down an estimate of the solution correct to 2 decimal places. (1 mark)
●●●
●●
Level 3
4 marks
Start
→
Mark as done
(a)
Show that the equation
x
3
−
5
x
+
1
=
0
x^3 - 5x + 1 = 0
x
3
−
5
x
+
1
=
0
has a solution between
x
=
2
x = 2
x
=
2
and
x
=
3
x = 3
x
=
3
(2 marks)
(b)
The equation
x
3
−
5
x
+
1
=
0
x^3 - 5x + 1 = 0
x
3
−
5
x
+
1
=
0
can be rearranged to give
x
=
5
x
−
1
3
x = \sqrt[3]{5x - 1}
x
=
3
5
x
−
1
Starting with
x
0
=
2
x_0 = 2
x
0
=
2
, use the iteration formula
x
n
+
1
=
5
x
n
−
1
3
x_{n+1} = \sqrt[3]{5x_n - 1}
x
n
+
1
=
3
5
x
n
−
1
three times to find an estimate for the solution.
Give your answer correct to 3 decimal places. (2 marks)
●●●
●●
Level 3
4 marks
Start
→
Mark as done
(a)
Show that the equation
x
3
+
x
−
7
=
0
x^3 + x - 7 = 0
x
3
+
x
−
7
=
0
has a solution between
x
=
1
x = 1
x
=
1
and
x
=
2
x = 2
x
=
2
. [2]
(b)
Show that the equation can be rearranged to
x
=
7
−
x
3
x = \sqrt[3]{7 - x}
x
=
3
7
−
x
. [1]
(c)
Starting with
x
0
=
1.5
x_0 = 1.5
x
0
=
1.5
, use the iteration
x
n
+
1
=
7
−
x
n
3
x_{n+1} = \sqrt[3]{7 - x_n}
x
n
+
1
=
3
7
−
x
n
to find
x
3
x_3
x
3
. Give your answer correct to 3 decimal places. [2]
●●●
●●
Level 3
5 marks
Start
→
Mark as done
x
n
+
1
=
7
x
n
+
8
3
,
x
1
=
3
x_{n+1} = \sqrt[3]{7x_n + 8}, \qquad x_1 = 3
x
n
+
1
=
3
7
x
n
+
8
,
x
1
=
3
Work out the values of
x
2
x_2
x
2
and
x
3
x_3
x
3
Give your answers correct to 3 decimal places.
●●●●
●
Level 4
3 marks
Start
→
Mark as done
x
n
+
1
=
20
−
x
n
3
,
x
1
=
2.5
x_{n+1} = \sqrt[3]{20 - x_n}, \qquad x_1 = 2.5
x
n
+
1
=
3
20
−
x
n
,
x
1
=
2.5
Work out the values of
x
2
x_2
x
2
and
x
3
x_3
x
3
Give your answers correct to 4 decimal places.
●●●●
●
Level 4
3 marks
Start
→
Mark as done
x
n
+
1
=
7
x
n
+
8
3
x_{n+1} = \sqrt[3]{7x_n + 8}
x
n
+
1
=
3
7
x
n
+
8
Given that
x
3
=
3.09009
x_3 = 3.09009
x
3
=
3.09009
, work out the value of
x
4
x_4
x
4
Give your answer correct to 3 decimal places.
●●●●
●
Level 4
2 marks
Start
→
Mark as done
x
n
+
1
=
3
x
n
+
7
,
x
1
=
4
x_{n+1} = \sqrt{3x_n + 7}, \qquad x_1 = 4
x
n
+
1
=
3
x
n
+
7
,
x
1
=
4
(a)
Work out the values of
x
2
x_2
x
2
and
x
3
x_3
x
3
, correct to 3 decimal places.
(b)
Explain how you could find a more accurate solution of
x
2
−
3
x
−
7
=
0
x^2 - 3x - 7 = 0
x
2
−
3
x
−
7
=
0
●●●●
●
Level 4
3 marks
Start
→
Mark as done
x
n
+
1
=
4
x
n
+
9
3
,
x
1
=
2
x_{n+1} = \sqrt[3]{4x_n + 9}, \qquad x_1 = 2
x
n
+
1
=
3
4
x
n
+
9
,
x
1
=
2
Work out the values of
x
2
x_2
x
2
and
x
3
x_3
x
3
Give your answers correct to 3 decimal places.
●●●●
●
Level 4
3 marks
Start
→
Mark as done
x
n
+
1
=
6
x
n
−
2
,
x
1
=
5
x_{n+1} = \sqrt{6x_n - 2}, \qquad x_1 = 5
x
n
+
1
=
6
x
n
−
2
,
x
1
=
5
Work out the values of
x
2
x_2
x
2
and
x
3
x_3
x
3
Give your answers correct to 4 significant figures.
●●●●
●
Level 4
3 marks
Start
→
Mark as done
x
n
+
1
=
x
n
3
+
3
8
,
x
1
=
0.4
x_{n+1} = \dfrac{x_n^3 + 3}{8}, \qquad x_1 = 0.4
x
n
+
1
=
8
x
n
3
+
3
,
x
1
=
0.4
Work out the values of
x
2
x_2
x
2
and
x
3
x_3
x
3
Give your answers correct to 3 decimal places.
●●●●
●
Level 4
3 marks
Start
→
Mark as done
(a)
Show that the equation
x
3
−
3
x
−
7
=
0
x^3 - 3x - 7 = 0
x
3
−
3
x
−
7
=
0
can be written in the form
x
=
3
x
+
7
3
x = \sqrt[3]{3x + 7}
x
=
3
3
x
+
7
(1 mark)
(b)
Starting with
x
0
=
2
x_0 = 2
x
0
=
2
use the iteration formula
x
n
+
1
=
3
x
n
+
7
3
x_{n+1} = \sqrt[3]{3x_n + 7}
x
n
+
1
=
3
3
x
n
+
7
three times to find an estimate for a solution of
x
3
−
3
x
−
7
=
0
x^3 - 3x - 7 = 0
x
3
−
3
x
−
7
=
0
(3 marks)
●●●●
●
Level 4
4 marks
Start
→
Mark as done
x
n
+
1
=
7
x
n
+
8
3
x_{n+1} = \sqrt[3]{7x_n + 8}
x
n
+
1
=
3
7
x
n
+
8
Starting with
x
1
=
3
x_1 = 3
x
1
=
3
, the values of
x
n
x_n
x
n
get closer and closer to
3.0958
…
3.0958\ldots
3.0958
…
(a)
Write down a solution, correct to 2 decimal places, of the equation
x
3
−
7
x
−
8
=
0
x^3 - 7x - 8 = 0
x
3
−
7
x
−
8
=
0
(b)
Explain why this value is a solution of the equation
x
3
−
7
x
−
8
=
0
x^3 - 7x - 8 = 0
x
3
−
7
x
−
8
=
0
●●●●●
Level 5
2 marks
Start
→
Mark as done
x
n
+
1
=
4
x
n
+
9
3
x_{n+1} = \sqrt[3]{4x_n + 9}
x
n
+
1
=
3
4
x
n
+
9
(a)
Given that
x
3
=
2.68168
x_3 = 2.68168
x
3
=
2.68168
, work out the value of
x
4
x_4
x
4
. Give your answer correct to 3 decimal places.
(b)
The values of
x
n
x_n
x
n
approach a limit. Which equation does that limit satisfy?
●●●●●
Level 5
4 marks
Start
→
Mark as done
x
n
+
1
=
30
−
2
x
n
3
,
x
1
=
3
x_{n+1} = \sqrt[3]{30 - 2x_n}, \qquad x_1 = 3
x
n
+
1
=
3
30
−
2
x
n
,
x
1
=
3
(a)
Work out the values of
x
2
x_2
x
2
and
x
3
x_3
x
3
. Give your answers correct to 4 decimal places.
(b)
Explain what the limit of this iteration represents.
●●●●●
Level 5
4 marks
Start
→
Rearranging an equation into iterative form
13 questions
Lesson
Not started
Mark as done
Show that the equation
x
3
−
7
x
−
8
=
0
x^3 - 7x - 8 = 0
x
3
−
7
x
−
8
=
0
can be rearranged to give
x
=
7
x
+
8
3
x = \sqrt[3]{7x + 8}
x
=
3
7
x
+
8
●●●
●●
Level 3
2 marks
Start
→
Mark as done
Show that the equation
x
3
+
x
−
20
=
0
x^3 + x - 20 = 0
x
3
+
x
−
20
=
0
can be rearranged to give
x
=
20
−
x
3
x = \sqrt[3]{20 - x}
x
=
3
20
−
x
●●●
●●
Level 3
2 marks
Start
→
Mark as done
Show that the equation
x
2
−
3
x
−
7
=
0
x^2 - 3x - 7 = 0
x
2
−
3
x
−
7
=
0
can be rearranged to give
x
=
3
x
+
7
x = \sqrt{3x + 7}
x
=
3
x
+
7
●●●
●●
Level 3
2 marks
Start
→
Mark as done
Show that the equation
x
3
−
4
x
−
9
=
0
x^3 - 4x - 9 = 0
x
3
−
4
x
−
9
=
0
can be rearranged to give
x
=
4
x
+
9
3
x = \sqrt[3]{4x + 9}
x
=
3
4
x
+
9
●●●
●●
Level 3
2 marks
Start
→
Mark as done
Show that the equation
x
2
−
6
x
+
2
=
0
x^2 - 6x + 2 = 0
x
2
−
6
x
+
2
=
0
can be rearranged to give
x
=
6
x
−
2
x = \sqrt{6x - 2}
x
=
6
x
−
2
●●●
●●
Level 3
2 marks
Start
→
Mark as done
(a)
Show that the equation
x
3
−
3
x
−
5
=
0
x^3 - 3x - 5 = 0
x
3
−
3
x
−
5
=
0
can be rearranged to give
x
=
3
x
+
5
3
x = \sqrt[3]{3x + 5}
x
=
3
3
x
+
5
(1 mark)
(b)
Starting with
x
0
=
2
x_0 = 2
x
0
=
2
, use the iteration
x
n
+
1
=
3
x
n
+
5
3
x_{n+1} = \sqrt[3]{3x_n + 5}
x
n
+
1
=
3
3
x
n
+
5
to find
x
1
x_1
x
1
,
x
2
x_2
x
2
and
x
3
x_3
x
3
. Give your answers correct to 4 decimal places. (2 marks)
(c)
Write down an estimate of the solution correct to 2 decimal places. (1 mark)
●●●
●●
Level 3
4 marks
Start
→
Mark as done
(a)
Show that the equation
x
3
+
x
−
7
=
0
x^3 + x - 7 = 0
x
3
+
x
−
7
=
0
has a solution between
x
=
1
x = 1
x
=
1
and
x
=
2
x = 2
x
=
2
. [2]
(b)
Show that the equation can be rearranged to
x
=
7
−
x
3
x = \sqrt[3]{7 - x}
x
=
3
7
−
x
. [1]
(c)
Starting with
x
0
=
1.5
x_0 = 1.5
x
0
=
1.5
, use the iteration
x
n
+
1
=
7
−
x
n
3
x_{n+1} = \sqrt[3]{7 - x_n}
x
n
+
1
=
3
7
−
x
n
to find
x
3
x_3
x
3
. Give your answer correct to 3 decimal places. [2]
●●●
●●
Level 3
5 marks
Start
→
Mark as done
An iteration for the equation
x
3
−
7
x
−
8
=
0
x^3 - 7x - 8 = 0
x
3
−
7
x
−
8
=
0
gives
x
4
=
3.094
x_4 = 3.094
x
4
=
3.094
Ella says, "
x
4
x_4
x
4
is the exact solution of the equation."
Explain why Ella is wrong, and state how a more accurate estimate could be found.
●●●●
●
Level 4
2 marks
Start
→
Mark as done
Show that the equation
x
3
+
2
x
−
30
=
0
x^3 + 2x - 30 = 0
x
3
+
2
x
−
30
=
0
can be rearranged to give
x
=
30
−
2
x
3
x = \sqrt[3]{30 - 2x}
x
=
3
30
−
2
x
●●●●
●
Level 4
2 marks
Start
→
Mark as done
Show that the equation
x
3
−
8
x
+
3
=
0
x^3 - 8x + 3 = 0
x
3
−
8
x
+
3
=
0
can be rearranged to give
x
=
x
3
+
3
8
x = \dfrac{x^3 + 3}{8}
x
=
8
x
3
+
3
●●●●
●
Level 4
2 marks
Start
→
Mark as done
(a)
Show that the equation
x
3
−
3
x
−
7
=
0
x^3 - 3x - 7 = 0
x
3
−
3
x
−
7
=
0
can be written in the form
x
=
3
x
+
7
3
x = \sqrt[3]{3x + 7}
x
=
3
3
x
+
7
(1 mark)
(b)
Starting with
x
0
=
2
x_0 = 2
x
0
=
2
use the iteration formula
x
n
+
1
=
3
x
n
+
7
3
x_{n+1} = \sqrt[3]{3x_n + 7}
x
n
+
1
=
3
3
x
n
+
7
three times to find an estimate for a solution of
x
3
−
3
x
−
7
=
0
x^3 - 3x - 7 = 0
x
3
−
3
x
−
7
=
0
(3 marks)
●●●●
●
Level 4
4 marks
Start
→
Mark as done
(a)
Show that the equation
x
3
−
5
x
+
1
=
0
x^3 - 5x + 1 = 0
x
3
−
5
x
+
1
=
0
can be rearranged to give
x
=
x
3
+
1
5
x = \dfrac{x^3 + 1}{5}
x
=
5
x
3
+
1
(b)
Using
x
n
+
1
=
x
n
3
+
1
5
x_{n+1} = \dfrac{x_n^3 + 1}{5}
x
n
+
1
=
5
x
n
3
+
1
with
x
1
=
0.2
x_1 = 0.2
x
1
=
0.2
, work out the value of
x
2
x_2
x
2
●●●●●
Level 5
3 marks
Start
→
Mark as done
(a)
Show that the equation
x
3
−
4
x
−
9
=
0
x^3 - 4x - 9 = 0
x
3
−
4
x
−
9
=
0
can be rearranged to give
x
=
x
3
−
9
4
x = \dfrac{x^3 - 9}{4}
x
=
4
x
3
−
9
(b)
Using the iteration
x
n
+
1
=
4
x
n
+
9
3
x_{n+1} = \sqrt[3]{4x_n + 9}
x
n
+
1
=
3
4
x
n
+
9
with
x
1
=
2
x_1 = 2
x
1
=
2
, work out
x
2
x_2
x
2
. Give your answer correct to 4 decimal places.
●●●●●
Level 5
4 marks
Start
→