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Grade 8 · Higher · free GCSE maths worksheet

Iterative methods for solving equations

Using x(n+1) = g(x(n)).

13 original questions, 52 marks. Print the worksheet, mark it with the answers sheet, and use the solutions to see every step written out by hand.

Worksheet PDFWorked solutions PDFAnswers PDF

The questions

1

f(x) = x3 − 4x − 3

Show that the equation f(x) = 0 has a root between x = 2 and x = 3.

2 marks
Show answer
f(2) = -3 and f(3) = 12, a change of sign
2

The number of members of a gym is modelled by the rule

un+1 = 1.2un − 50

where un is the number of members after n years.

There are 500 members now, so u0 = 500.

Work out the number of members the model predicts after 2 years.

2 marks
Show answer
610
3

A ball is dropped from a height of 5 m onto a hard floor.

After each bounce the ball rises to 60% of the height of the previous bounce, so

hn+1 = 0.6hn, where h0 = 5

  1. Work out the height the ball rises to after the third bounce.

  2. After which bounce does the ball first rise to less than 1 m?

3 marks
Show answer
a) 1.08 m b) 4th bounce
4

Amara opens a savings account with £1000.

The account pays 3% interest at the end of each year, and Amara then pays in another £200.

The amount in the account after n years is given by

an+1 = 1.03an + 200, where a0 = 1000

Work out the amount in the account after 3 years. Give your answer to the nearest penny.

3 marks
Show answer
£1710.91
5

A sequence is given by the iteration formula

xn+1 = 3√(5xn + 2), with x1 = 2

  1. Work out the values of x2, x3 and x4. Give each value correct to 3 decimal places.

  2. The values of xn approach a root of an equation. Write down this equation in the form x3 + ax + b = 0

4 marks
Show answer
a) x2 = 2.289, x3 = 2.378, x4 = 2.404 b) x3 - 5x - 2 = 0
6

A sequence is defined by

un+1 = kun + 3

where k is a constant.

u1 = 5 and u2 = 13

  1. Find the value of k

  2. Find the value of u4

4 marks
Show answer
a) k = 2 b) 61
7

The number of deer in a woodland is modelled by

pn+1 = 0.9pn + 40

where pn is the number of deer n years from now.

There are 300 deer now, so p0 = 300.

  1. Work out the number of deer the model predicts in 2 years' time.

  2. The model predicts that in the long term the number of deer approaches a fixed value L. Work out the value of L.

4 marks
Show answer
a) 319 b) 400
8
  1. Show that the equation x3 − 7x + 3 = 0 can be rearranged to give

    x = 3√(7x − 3)

  2. Using the iteration formula xn+1 = 3√(7xn − 3) with x1 = 2, work out the values of x2, x3 and x4. Give each value correct to 3 decimal places.

4 marks
Show answer
a) x3 = 7x - 3, so x = cube root of (7x - 3) b) x2 = 2.224, x3 = 2.325, x4 = 2.368
9

f(x) = x2 − 6x + 2

  1. Show that the equation f(x) = 0 has a root between x = 0 and x = 1

  2. Show that the equation x2 − 6x + 2 = 0 can be rearranged to give

    x = x2 + 26

4 marks
Show answer
a) f(0) = 2 > 0 and f(1) = -3 < 0, sign change so a root lies between 0 and 1 b) 6x = x2 + 2 so x = (x2 + 2)/6
10

f(x) = x3 + 4x − 9

  1. Show that the equation f(x) = 0 has a root between x = 1 and x = 2

  2. Show that the equation x3 + 4x − 9 = 0 can be rearranged to give

    x = 3√(9 − 4x)

  3. Starting with x0 = 1.5, use the iteration formula xn+1 = 3√(9 − 4xn) to work out the values of x1, x2 and x3. Give each value correct to 4 decimal places.

6 marks
Show answer
a) f(1) = -4 and f(2) = 7; the sign change means a root lies between 1 and 2 b) x3 = 9 - 4x, so x = cube root of (9 - 4x) c) x1 = 1.4423, x2 = 1.4784, x3 = 1.4560
11

f(x) = x2 + 2x − 5

  1. Show that the equation f(x) = 0 has a root between x = 1 and x = 2

  2. Show that the equation x2 + 2x − 5 = 0 can be rearranged to give

    x = 5x + 2

  3. Starting with x0 = 1.5, use the iteration formula xn+1 = 5xn + 2 to work out the values of x1, x2 and x3. Give each value correct to 4 decimal places.

  4. Write down an estimate of this root of f(x) = 0 correct to 2 decimal places.

6 marks
Show answer
a) f(1) = -2 and f(2) = 3; the sign change means a root lies between 1 and 2 b) x2 + 2x = 5, so x(x + 2) = 5 and x = 5/(x + 2) c) x1 = 1.4286, x2 = 1.4583, x3 = 1.4458 d) 1.45
12

The equation x3 − 4x − 3 = 0 has a root between 2 and 3.

Use the iteration formula xn+1 = 3√(4xn + 3) with x0 = 2.5.

  1. Work out x1, x2 and x3. Give each answer to 4 decimal places.
  2. By evaluating f(x) = x3 − 4x − 3 at 2.295 and 2.305, show that the root is 2.30 correct to 2 decimal places.
5 marks
Show answer
a) x1 = 2.3513, x2 = 2.3149, x3 = 2.3058 b) f(2.295) = -0.092, f(2.305) = 0.027, root = 2.30
13

x3 − 2x2 − 1 = 0 has a root between 2 and 3.

  1. Show that the equation can be rearranged to give x = 2 + 1x2.
  2. Use the iteration xn+1 = 2 + 1xn2 with x0 = 2 to find x1, x2 and x3, to 4 decimal places.
  3. Given that f(2.205) = −0.0033 and f(2.215) = 0.0548 where f(x) = x3 − 2x2 − 1, give the root to 2 decimal places.
5 marks
Show answer
a) divide x3 = 2x2 + 1 by x2 b) x1 = 2.25, x2 = 2.1975, x3 = 2.2071 c) 2.21

Every question here is original and written by Robinson Tuition. The grade follows the Edexcel grade descriptors and suits AQA and OCR too; this is not an exam-board paper.

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