Video Transcript
Use the method of trial and improvement to find the positive solution of the equation 104 over 𝑥 squared minus three equals 27, given that it lies between one and two. Give your answer to two decimal places.
In this question, we’re asked to solve or find the solution of the equation of 104 over 𝑥 squared minus three equals 27. But rather than solving this through rearranging or balancing, we’re asked to use trial and improvement. That means we’re going to take different values of 𝑥 and plug them into this expression, 104 over 𝑥 squared minus three, and see how close we get to an answer of 27. We can start with any value for 𝑥, for example, 100 or negative five. But in this, we’re given the clue that 𝑥 is between one and two, so we’ll start with these values.
The best way to set out our workings for trial and improvement is to use a table. The first column in our table will be the place where we record the value of 𝑥 that we’re trialing. We need a column to record the value of our expression, 104 over 𝑥 squared minus three, with our value of 𝑥. We might also find it helpful to break this expression down into a smaller part, for example, just calculating 104 over 𝑥 squared in order to help us with our calculation. The final column of the table is where we record if the value that we found is too big or too small.
The first two values that we’re going to try are the values of 𝑥 equals one and 𝑥 equals two. This will give us a better understanding of the next values of 𝑥 that we’ll choose after that. We’ll be using a calculator throughout this question. So, in our first row, we take our value of 𝑥 equals one and plug it into 104 over 𝑥 squared. As this is the same as 104 over one, then our value is 104. This value subtract three gives us 101. Remembering that we’re comparing 101 with 27, we can say that this result is too big. Next, in our second row, we’re calculating 104 over two squared, which gives us the value of 26. In our next column then, subtracting three from that would give us 23. Comparing this with the value of 27, we can say that this is too small.
Let’s pause and take a minute to look at the results of these two trials. How come whenever 𝑥 is one, we got a value that was too big, and when we made it larger to 𝑥 equals two, the result was too small? Well, it’s all to do with this part of the expression, 104 over 𝑥 squared. Whenever we take a larger value of 𝑥 and square it, then this whole expression, 104 over 𝑥 squared, actually gets smaller. Usually, when we’re doing a trial and improvement question, if we want the result to be larger, we take a larger value of 𝑥. The opposite is true in this question. If our result is small and we want it to be a larger value, we need to take a smaller value of 𝑥 in the next trial.
So, which value of 𝑥 shall we choose for our next trial? We know that it’s got to be between one and two, and it should be smaller than two. So we could look at the results 101 and 23 to help us decide. The value of 23 is closer to 27 than the value of 101, so our 𝑥-value should be closer to two than one. Let’s try 𝑥 is equal to 1.9. Plugging in this value into our expression gives us the result of 25.8088. We don’t need to worry about recording all the decimal places. It’s just enough to give us an indication. This value is too small. It’s smaller than 27. So, remember that this means that our next value will be smaller than 1.9.
Let’s find the value when 𝑥 is equal to 1.8. The result of 29.0987 and so on is too big as it’s larger than 27. At this point, we can note that we have narrowed down the range of the values of 𝑥 between these last two trials. At the start, we were told that 𝑥 lies between one and two, and now we can say that 𝑥 lies between 1.8 and 1.9. For the next trial of 𝑥 then, we can notice that the value of 27 lies almost in between our two values of 25.80 and 29.09. It may be slightly closer to 1.9, but it doesn’t really matter which value we choose. So let’s try 𝑥 is equal to 1.86. The result here will be a value that’s too big. So our next trial, remember, we’ll choose a larger value, so let’s try 1.87 for 𝑥. This result of 26.7406 and so on is too small.
However, once again, if we look at these two trials, we find that the value must lie between 1.86 and 1.87. We’re asked to give our answer to two decimal places. So, which value of 𝑥 should we say, 1.86 or 1.87? The answer is that we choose the value in between, 1.865, and do one last trial with 𝑥 is equal to 1.865. The result of this is 26.9003 which is too small as it’s smaller than 27. Because of the nature of this question and the fact that when 𝑥 is larger, our result of 104 over 𝑥 squared minus three gets smaller, then how we choose the value of 𝑥 will be different than in most questions. Let’s go back to this little sketch of the range of our values of 𝑥.
When 𝑥 was equal to 1.86, the result was 27.06 and so on. And when 𝑥 was equal to 1.87, the result was 26.74 and so on. Our test trial when 𝑥 was equal to 1.865 gave us a result of 26.90 and so on. So, where would our ideal result of 27 actually lie? Well, it would be in this range between 1.86 and 1.865. If we were to choose any of the values within this range and round them to two decimal places, we’d get the value of 1.86. And so, that’s our solution to the equation 104 over 𝑥 squared minus three equals 27 to two decimal places.
