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Question Video: Using Trial and Improvement to Find a Solution to Two Decimal Places Mathematics

By using a trial and improvement method, find the positive solution of 𝑦² + 𝑦 = 1000 to two decimal places, given that it lies between 31 and 32.

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Video Transcript

By using a trial and improvement method, find the positive solution of 𝑦 squared plus 𝑦 equals 1000 to two decimal places, given that it lies between 31 and 32.

Here we have this equation, 𝑦 squared plus 𝑦 equals 1000. And we’re asked to find the solution to it, which means we’re really trying to figure out what the value of 𝑦 is. In this question, however, we’re not going to try and solve it using an algebraic method, but instead by trial and improvement. In a trial and improvement method, we take different values here of 𝑦, plugging them into the equation 𝑦 squared plus 𝑦, to see if we can get a value of 1000. Very often we find we’ll not get exactly 1000, but we’re trying to get as close as possible.

The best way to set out the workings for trial and improvement is by using a table. We can set up a table. In the first column, we need a place to note the value of 𝑦 that we’re choosing. We’ll also need to record the value when we have 𝑦 squared plus 𝑦. I’ve also added another column here in order to break this expression down into more manageable chunks, but it’s not always necessary. The final column is the place where we can record if the value of 𝑦 that we’ve chosen was too big or too small.

So let’s get started. Now, we could choose any value of 𝑦, but here we were given the hint that the value of 𝑦 is between 31 and 32. The first two trials we can do then is taking 𝑦 is equal to 31 and 𝑦 is equal to 32. This will help to give us an indication of how close the value of 𝑦 will be to 31 or to 32. Using a calculator throughout this question, we can start then and find the value of 31 squared, which is 961. Adding 31 and 961 gives us the value of 992. Comparing this value of 992 with the value of 1000 tells us that the value of 31 that we chose is too small.

In the second trial then, we square 32 which gives us 1024. Adding 32 and 1024 gives us the value of 1056. This value is larger than 1000, so we know that this value is too large. We already knew this anyway because we were told that the solution lies between 31 and 32. So which value will we choose next? Well, if we look at our results, 992 is closer to 1000 because it’s only eight away, whereas 1056 is 56 away. Choosing a value like 31.1 might be good as it’s closer to 31. We can then evaluate 𝑦 squared plus 𝑦 using the value of 𝑦 equals 31.1 to give us the value of 998.31.

Even though we’re getting closer to 1000, this value will still be too small. As it’s not too far away from 1000, we’re not going to choose a much larger value. Say something like 31.2 would be a sensible trial for the next one. Evaluating 31.2 plus 31.2 squared gives us the value of 1004.64. This value is too large, but it has helped us to narrow down our search. We were told in the question that the value of 𝑦 is between 31 and 32. And now we’ve established that 𝑦 is between 31.1 and 31.2.

We’re asked to find the value of 𝑦 to two decimal places. So we need to narrow this range even further. So which value shall we try next? Well, if we look at our two results for 31.1 and 31.2, we can see that the result 998.31 is less than two away from 1000, whereas the value for 31.2 is more than four away. We could guess, therefore, that the next value should be closer to 31.1 rather than 31.2. When we try the value of 𝑦 as 31.11, our result is too small, so we need to try a larger value.

When 𝑦 is 31.12, the result was too small. So when we try a larger value of 31.13, we got a result that was too large. We have however shortened the range of possible values that 𝑦 could be. We know that it must lie between 31.12 and 31.13. We’re asked to give a value of 𝑦 to two decimal places. But how do we know which of these values it would be? We do this by testing the value in the middle.

Our final trial then will be when 𝑦 is equal to 31.125. When we plug in this value into 𝑦 squared plus 𝑦, we get 999.8906 and continuing, which is too small as it’s less than 1000. So how does this help us find the value to two decimal places? Well, once again, we have narrowed the range of possible values for 𝑦. We’ve shown that it must be between 31.125 and 31.13. And if we were to take any of the values in this range and round them to two decimal places, we’d get the answer 31.13.

We can then give our answer that the solution of 𝑦 squared plus 𝑦 equals 1000 to two decimal places is 31.13.

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