YOU ARE LEARNING:
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=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 x. However, the value of x may not ever be exact.
What is an iterative formula?
An iterative formula is in the form x=, and allows us to work out values of x.
Let's try making an iterative formula for x3+x=5
Rearrange to make x the subject
We need to find an equation with x as the subject in its singular form. Since the two x terms are raised to different powers, we do not collect them together.
First, make x3 the subject
Then cube root both sides to isolate singularx
x=3(5−x) is our iterative formula which we can use to find increasingly accurate values of x.
Let's try another one! Make an iterative formula for x2+x=56
Make x2 the subject of the equation.
Square root both sides
Write an iterative formula for x2−6x=7
Write an iterative formula for x2+5x−24=0
Once we have an iterative formula, we can use it to find values for x.
We will call the x on the left of the formula Xn+1 and the x on the right Xn.
This just means that we will start with a value of X1 on the right and use the formula to find the second value, X2. We then repeat this process to find X3, and so on.
Let's use the iterative formula to find some values :Xn+1=35−Xn
The first value of n is 1
It is the first time using the formula
We can choose any value to try for X1
We will use 2
Put these numbers into the formula
Find the value of X2
X2=35−2 so X2=1.44224957
Use X2 to find X3
Solve, to 4 s.f., X2+1=35−1.44224957
X3=1.526599655. What is X4 if X3+1=35−1.526599655
X4=1.514438405. What is X5 if X4+1=35−1.514438405
Notice the difference between X values
The difference between successive values is getting smaller. This means that we are getting more accurate values.
So, we can round our number
X5 and values after it all round to 1.52
This is our answer!
To solve x+1−x=0, we can use the iterative formula: Xn+1=Xn+1. If X1=2, what is X2?
To solve x+1−x=0, we can use the iterative formula: Xn+1=Xn+1. If X2=2.414213562 what is X3?
If you keep iterating using the formula Xn+1=Xn+1, what is the approximate solution to x+1−x=0 to 3 significant figures?