Video: Using Trial and Improvement to Find a Solution to One Decimal Place

Use a trial and improvement method to solve the equation 𝑥³ + 2𝑥² − 25 = 0 to one decimal place, knowing that the equation has a solution between 2 and 3.

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Video Transcript

Use a trial-and-improvement method to solve the equation 𝑥 cubed plus two 𝑥 squared minus 25 equal zero to one decimal place, knowing that the equation has a solution between two and three.

In this question, we’re asked to solve the equation. This means that we need to find the value of 𝑥 for which 𝑥 to the third power plus two 𝑥 squared minus 25 would equal zero. We can solve equations in a number of ways, for example, by factoring or graphically. However, here we’re asked to solve using trial and improvement. This means that we’re going to try different values of 𝑥 to see if we can get our equation close to zero.

We can start by using any values of 𝑥. For example, 𝑥 is negative 10 or 𝑥 is 100. But we’re actually given a starting point that 𝑥 is between two and three. So we’re going to start with those values. The important thing with the trial-and-improvement method is finding a neat way to present our workings. And we can do this using a table.

We start with a column for the value of 𝑥 that we’re going to be using. For example, the first trial will be 𝑥 equals two. We need a column for calculating 𝑥 to the third power plus two 𝑥 squared minus 25 for our value of 𝑥. And it may be helpful to add three columns in between those two to break down the equation into smaller pieces. And finally, we add a column to record whether the value of our equation is too big or too small in comparison to zero.

So let’s then start with our first trial of 𝑥 equals two, since we were told that the equation has a solution between two and three. We substitute 𝑥 equals two into our equation. So that means that every value of 𝑥 here will be replaced with the value two. We can recall that two cubed is the same as two times two times two, which is eight. When we have two times two squared, this is the same as two times two times two, which is also equal to eight.

And next, the value of negative 25 will always stay the same. So to work out the value of our equation, we take the three pieces of information on 𝑥 to the third power, two 𝑥 squared, and negative 25, which is eight plus eight take away 25. So the value of our equation will be negative nine. We could also have put the whole equation with 𝑥 equals two into our calculator, which would’ve given us the same value of negative nine.

We now need to check if our value of negative nine is bigger or smaller than zero. And since it’s less than zero, we know that our value of 𝑥 must be too small. So for our next trial, we’re going to use the value of 𝑥 equals three. So then three to the third power will be three times three times three, which is 27. Next, two times three squared will be the same as two times three times three, giving us 18. And our negative 25 stays the same. Putting it together then, we have 27 plus 18 minus 25, which will give us an answer of 20.

Next, how is our value of 20 compared with the value of zero? Well, it’s too big. And note that this does also confirm that our equation does have a solution between two and three. So now let’s try a value of 𝑥 that’s smaller than three. We could use the value of 𝑥 equals 2.5 since that’s nicely in between two and three. We can start by calculating 2.5 to the third power, which we can evaluate as 15.625. Next, two times 2.5 squared will give us 12.5. And our negative 25 stays the same. Putting it together, we have 15.625 plus 12.5 minus 25, which is equal to 3.125. And then since 3.125 is bigger than zero, our value of 𝑥 must be too big.

This means, for our next value of 𝑥, we should choose a number that’s less than 2.5. Let’s try 𝑥 equals 2.4. At this point, we can add a handy calculator tip, particularly if you’ve been working out the whole of the equation in one go. If your calculator has a replay function, this means that you can change the values that were last entered to a different value. In this case, you could change all the values that said 2.5 into 2.4.

For each individual term, when 𝑥 equals 2.4, we would have 13.824, 11.52, and negative 25. The value of our equation will be 0.344. And since this is larger than zero, then our 𝑥-value is too big. We can notice, however, that our equation is quite close to zero. So our next smaller value of 𝑥 shouldn’t be too much smaller than 2.4. So let’s choose 𝑥 equals 2.3.

Carrying out our calculations will give us 𝑥 to the third power plus two 𝑥 squared minus 25 as negative 2.253. So this means that the value of 𝑥 is too small. However, we can notice that we have now narrowed it down to one decimal place. Since we have 𝑥 equals 2.3 is too small and 𝑥 equals 2.4 is too big, then our value of 𝑥 must be somewhere in between these values.

We were asked for an answer to one decimal place. So how do we choose between 2.3 and 2.4? And the answer is that we choose the value in between these two to check. And since that value is 2.35, then we carry out one more trial with 𝑥 equals 2.35 and check the result. And the value of our equation with 𝑥 equals 2.35 will be negative 0.977125. The important thing here is that this value is too small.

If we visualise this on a number line, we can now see that 𝑥 must be between 2.35 and 2.4. If we took any of the values in this range and write them down to one decimal place, we would get 2.4. This means that our solution to one decimal place is 2.4.

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