Lesson video

Hello, my name is Mr Clasper and today we are going to be using trial and improvement to find approximate solutions to equations.

How would you show there is a solution to the equation below between the values where x is equal to two and where x is equal to three? Well, if we substitute both of these values, we return a value of 20 when we substitute two and we return a value of 54 when we substitute the value of three.

As our aim is to find a value of 30, which is between the values of 20 and 54, we must have a solution between where x is equal to two and where x is equal to three.

How can you find the solution to a greater degree of accuracy? How could you find a solution to one decimal place? Well, if I tried the midpoint of two and three which is 2.

5, this returns a value of 34.

375.

However, this is too big as I'm aiming for the number 30.

So, let's try a smaller number.

If I substitute 2.

4, I return a value of 31.

104.

This is too big again as I need a value of 30.

Let's try a smaller number.

Let's try 2.

3.

This returns a value of 28.

037.

This is now too small.

This means I must have a solution between 2.

3 and 2.

4, but how do I decide which of these two is my correct solution? Well, if I try the midpoint of these two numbers, I return a value of 29.

545 and this is too small.

Looking at this diagram.

We can see that if 2.

35 would return a value of 29.

545, the only way I can return a value closer to 30 is if I choose a number which is greater than 2.

35, which means that my solution to one decimal place would have to be 2.

4.

Here's a question for you to try.

Pause the video to complete your task and click resume once you're finished.

And here is your solution.

So, if we look at our table, once we tried 3.

5 and established it was too small, we then tried 3.

6.

This was still too small and then we tried 3.

7 which was too big.

This verified that our solution must be between 3.

6 and 3.

7.

Now, the next thing to do is to try 3.

65 which is the mid point of 3.

6 and 3.

7 and when we do this, we find that this value is too small.

Therefore our solution must be closer to 3.

7 than it is to 3.

6.

Here's another question for you to try.

Pause the video to complete your task and click resume once you're finished.

And here is your solution.

So, once again, once we found the two numbers to one decimal place, which our solution resides between, if we tried the midpoint, which in this case was 5.

15, this was too big.

Therefore this means that our solution is closer to 5.

1 than it is to 5.

2.

And that brings us to the end of our lesson.

So, we've been using trial and improvement to find approximate solutions to equations.

Have a go at the exit quiz to show off your brand new skills.

I'll hopefully see you soon.