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Iteration is another method of find approximate solutions to complicated quadratic equations.

Iteration is used to find an approximate numerical solution to an equation. This is particularly useful for equations that are complicated to solve or do not have whole number solutions.

Look at x3+x=5x^3 + x = 5.

We don't have an easy algebraic way to solve this, but we can repeatedly use an iterative formula to find more accurate values of xx. However, the value of xx may not ever be exact.

What is an iterative formula?

An iterative formula is in the form x=x=, and allows us to work out values of xx.

Let's try making an iterative formula for x3+x=5x^3 + x = 5


Rearrange to make xx the subject

We need to find an equation with xx as the subject in its singular form. Since the two xx terms are raised to different powers, we do not collect them together.


First, make x3x^3 the subject

x3=5xx^3 = 5 - x


Then cube root both sides to isolate singularxx

x=(35x)x =\sqrt[3](5 - x)



x=(35x)x =\sqrt[3](5 - x) is our iterative formula which we can use to find increasingly accurate values of xx.

Let's try another one! Make an iterative formula for x2+x=56x^2 + x = 56


Make x2x^2 the subject of the equation.


Square root both sides

Write an iterative formula for x26x=7x^2 - 6x = 7

Write an iterative formula for x2+5x24=0x^2 + 5x -24 = 0

Once we have an iterative formula, we can use it to find values for xx.

We will call the xx on the left of the formula Xn+1X_{n+1} and the xx on the right XnX_n.

x=5x3x = \sqrt[3]{5 - x} \rightarrow

Xn+1=5Xn3X_{n+1} = \sqrt[3]{5 - X_n}

This just means that we will start with a value of X1X1 on the right and use the formula to find the second value, X2X2. We then repeat this process to find X3X3, and so on.

Let's use the iterative formula to find some values :Xn+1=5Xn3X_{n+1} = \sqrt[3]{5 - X_n}


The first value of nn is 1

It is the first time using the formula


We can choose any value to try for X1X_1

We will use 22


Put these numbers into the formula

X1+1=523X_{{\color{#21affb}1}+1} = \sqrt[3]{5 - {\color{#21affb}2}}


Find the value of X2X_2

X2=523X_2=\sqrt[3]{5 - 2} so X2=1.44224957X_2=1.44224957


Use X2X_2 to find X3X_3

X2+1=51.442249573X_{\color{#21affb}2}+1 = \sqrt[3]{5 - {\color{#21affb}1.44224957}}


Solve, to 4 s.f., X2+1=51.442249573X_{{\color{#21affb}2}+1} = \sqrt[3]{5 - {\color{#21affb}1.44224957}}


X3=1.526599655X3=1.526599655. What is X4X4 if X3+1=51.5265996553X_{{\color{#21affb}3}+1} = \sqrt[3]{5 - {\color{#21affb}1.526599655}}


X4=1.514438405X4=1.514438405. What is X5X5 if X4+1=51.5144384053X_{{\color{#21affb}4}+1} = \sqrt[3]{5 - {\color{#21affb}1.514438405}}


Notice the difference between XX values

The difference between successive values is getting smaller. This means that we are getting more accurate values.


So, we can round our number

X5X_5 and values after it all round to 1.521.52


This is our answer!

Approximately, x1.52x\approx1.52

To solve x+1x=0\sqrt{x} + 1 - x = 0, we can use the iterative formula: Xn+1=Xn+1X_n+1 = \sqrt{X_n} + 1. If X1=2X1 = 2, what is X2X2?

To solve x+1x=0\sqrt{x} + 1 - x = 0, we can use the iterative formula: Xn+1=Xn+1X_n+1 = \sqrt{X_n} + 1. If X2=2.414213562X_2 = 2.414213562 what is X3X_3?

If you keep iterating using the formula Xn+1=Xn+1X_n+1 = \sqrt{X_n} + 1, what is the approximate solution to x+1x=0\sqrt{x} + 1 - x = 0 to 3 significant figures?