x 3 – 4x + 1 = 0 x 3 – 9x + 1 = 0 x 2 – 5x + 1 = 0 x 2

Iterative processes
Iterative methods 1
Cut out the following equations and match with the iterative process.
Each equation matches two iterative processes.
x3 – 4x + 1 = 0
x3 – 9x + 1 = 0
x2 – 5x + 1 = 0
x2 – 4x + 1 = 0
x3 – 5x = -1
x2 – 9x + 1 = 0
xn+1 =
𝑥𝑛3 +1
xn+1 =
4
𝑥𝑛2 +1
4
xn+1 = 3√4𝑥𝑛 − 1
xn+1 = 3√5𝑥𝑛 − 1
xn+1 = 19 (𝑥𝑛2
xn+1 = 15 (𝑥𝑛3
+ 1)
+ 1)
xn+1 = √4𝑥𝑛 − 1
xn+1 = √9𝑥𝑛 − 1
xn+1 = 3√9𝑥𝑛 − 1
xn+1 = √5𝑥𝑛 − 1
xn+1 = 19 (𝑥𝑛3
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+ 1)
xn+1 =
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𝑥𝑛2 +1
5
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Iterative processes
Solutions
𝑥𝑛3 +1
x3 – 4x + 1 = 0
xn+1 = 3√4𝑥𝑛 − 1
xn+1 =
x3 – 9x + 1 = 0
xn+1 = 3√9𝑥𝑛 − 1
xn+1 = 19 (𝑥𝑛3
x2 – 5x + 1 = 0
xn+1 = √5𝑥𝑛 − 1
xn+1 =
x2 – 4x + 1 = 0
xn+1 = √4𝑥𝑛 − 1
xn+1 =
x3 – 5x = -1
xn+1 = 3√5𝑥𝑛 − 1
xn+1 = 15 (𝑥𝑛3
+ 1)
x2 – 9x + 1 = 0
xn+1 = √9𝑥𝑛 − 1
xn+1 = 19 (𝑥𝑛2
+ 1)
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4
+ 1)
𝑥𝑛2 +1
5
𝑥𝑛2 +1
4
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Iterative processes
Iterative methods 2
1. Rearrange each of the following equations to give an iterative formula.
a. x2 – 5x + 3 = 0
b. x2 + 8x + 5 = 0
c. x2 – x - 1 = 0
d. x2 + 2x = 5
e. x2 + 5x - 1 = 0
2. a. Show that 𝑥 2 + 𝑥 − 28 = 0 can be rearranged to give
𝑥 = √(28 − 𝑥)
b. Write this equation as the iterative formula
c. Using this iterative formula with x1 = 4, find an approximate solution to the
equation 𝑥 2 + 𝑥 − 28 = 0 to 2 decimal places
3. a) Show that 𝑥 3 − 6𝑥 = 120
can be rearranged to give
3
𝑥 = √6𝑥 + 120
b. Write this equation as the iterative formula
c. Using this iterative formula with x1 = 5, find an approximate solution to the
equation 𝑥 3 − 6𝑥 = 120 to 3 significant figures
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Iterative processes
Solutions
1.
a. x2 – 5x + 3 = 0
𝑥𝑛+1 = √5𝑥𝑛 − 3
𝑥𝑛+1 =
𝑥𝑛2 +3
5
𝑥𝑛2 +5
b. x2 - 8x + 5 = 0
𝑥𝑛+1 = √8𝑥𝑛 − 5
𝑥𝑛+1 =
c. x2 – x - 1 = 0
𝑥𝑛+1 = √𝑥𝑛 − 1
𝑥𝑛+1 = 𝑥𝑛2 + 1
d. x2 + 2x = 5
𝑥𝑛+1 = √5 − 2𝑥𝑛
𝑥𝑛+1 =
e. x2 + 5x - 1 = 0
𝑥𝑛+1 = √1 − 5𝑥𝑛
𝑥𝑛+1 =
8
5− 𝑥𝑛2
2
1− 𝑥𝑛2
5
There are other possible answers.
2.
b. 𝑥𝑛+1 = √(28 − 𝑥𝑛 )
c. 4.82
3.
b. 𝑥𝑛+1 = 3√6𝑥𝑛 + 120
c. 5.34
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