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# Trial And Improvement

## Introduction

An exam question that asks you to find a solution to an equation using the technique of trial and improvement is one of those classes of questions that can stump the best of maths students. In my experience the problem is not so much that the student has a problem understanding the process. It is that you have to have a feel for the numbers and that (trying not to sound too much like an old-timer ) only comes with lots of experience and practice.

I have seen trial and improvement taught in a variety of ways (and most recently taught completely wrongly - which is what has prompted me to write this short article) but the method I have outlined here is probably the best. The key thing is that you have got to be methodical!

## A Typical Question

Here is a typical trial and improvement question…

Find a solution to the following equation using trial and improvement, giving your answer to 1 dp

Remember: the key to answering questions of this sort is to be methodical!

We are going to be testing the equation with various values of hoping to home in on the correct answer to the question. The best way of recording our test results (like the good scientist I am ) is in a table, and it is best to draw up a table like so…

x x3 3x x3+3x Comment

Now we have to choose a value of x to start with. That will be our stake in the ground. For example, let's substitute into the original equation to see what we end up with.

Lets populate the table…

and

so

So when [itex]x[/itex] is 1 we get the answer 4, not 10. That answer is too low…

x x3 3x x3+3x Comment
1 1 3 4 Too Low

Because 1 was too low let's try [itex]x = 2[/itex]…

and

so

So when is 2 we get the answer 14, not 10. That answer is too high…

x x3 3x x3+3x Comment
1 1 3 4 Too low
2 8 6 1 Too high

Clearly the answer is a number between 1 and 2. Let's now see if we can find the next decimal place. Let's, for the sake of argument, try 1.5…

and

so

This is still too low…

x x3 3x x3+3x Comment
1 1 3 4 Too low
2 8 6 1 Too high
1.5 3.375 4.5 7.875 Too low

…so let's try 1.7

and

so

Too high!

## Introduction

An exam question that asks you to find a solution to an equation using the technique of trial and improvement is one of those classes of questions that can stump the best of maths students. In my experience the problem is not so much that the student has a problem understanding the process. It is that you have to have a feel for the numbers and that (trying not to sound too much like an old-timer ) only comes with lots of experience and practice.

I have seen trial and improvement taught in a variety of ways (and most recently taught completely wrongly - which is what has prompted me to write this short article) but the method I have outlined here is probably the best. The key thing is that you have got to be methodical!

## A Typical Question

Here is a typical trial and improvement question…

Find a solution to the following equation using trial and improvement, giving your answer to 1 dp

Remember: the key to answering questions of this sort is to be methodical!

We are going to be testing the equation with various values of hoping to home in on the correct answer to the question. The best way of recording our test results (like the good scientist I am ) is in a table, and it is best to draw up a table like so…

x x3 3x x3+3x Comment

Now we have to choose a value of x to start with. That will be our stake in the ground. For example, let's substitute into the original equation to see what we end up with.

Lets populate the table…

and

so

So when is 1 we get the answer 4, not 10. That answer is too low…

x x3 3x x3+3x Comment
1 1 3 4 Too Low

Because 1 was too low let's try

and

so

So when is 2 we get the answer 14, not 10. That answer is too high…

x x3 3x x3+3x Comment
1 1 3 4 Too low
2 8 6 1 Too high

Clearly the answer is a number between 1 and 2. Let's now see if we can find the next decimal place. Let's, for the sake of argument, try 1.5…

and

so

This is still too low…

x x3 3x x3+3x Comment
1 1 3 4 Too low
2 8 6 1 Too high
1.5 3.375 4.5 7.875 Too low
1.7 4.913 5.1 10.013 Too high

Let's try . I'll just write the numbers in the table without going through the maths - I guess by now you'll be getting the idea

x x3 3x x3+3x Comment
1 1 3 4 Too low
2 8 6 1 Too high
1.5 3.375 4.5 7.875 Too low
1.7 4.913 5.1 10.013 Too high
1.6 4.096 4.8 8.896 Too low

So the answer is somewhere between 1.6 and 1.7. And at this stage we have to put our thinking caps on because the question wants the answer to 1dp, but without checking the next decimal place it is impossible for us to know if the correct answer will be 1.6 or 1.7. Let's check 1.65…

x x3 3x x3+3x Comment
1 1 3 4 Too low
2 8 6 1 Too high
1.5 3.375 4.5 7.875 Too low
1.7 4.913 5.1 10.013 Too high
1.6 4.096 4.8 8.896 Too low
1.65 4.492125 4.95 9.442125 Too low

Now we actually have enough information to be able to answer this question: because 1.65 (which to 1dp is 1.7) is two low and 1.7 is too high then to 1dp must be 1.7 (i.e. is between 1.65 and 1.7. And there is our answer…

to 1dp