Trail and Improvement

In this chapter we shall be exploring the trail and improvement method. Trail and improvement is used to find approximate answers to quite complicated equations especially equations involving cubic (x³).
The trial and improvement method works by trying different values into an equation to find the one that gives the answer on one side, for example having to find the value of x in these equations; x³+x=40, x³-2x=1, x³+3x=100 etc. Usually they provide limit intervals for which the value of x must not extend, if you’re in an exam situation that is.
The following example shows how to work with the trail and improvement methods. The equation x³+x=40 has solution between 3 and 3.5. Find this solution to 1dp.
Here we’re asked to find the value of x that we could replace into the equation above to give 40. We’ve also been given a clue that this value is between 3 and 3.5. The value we find must be written in one decimal place.
Trail and improvement method works by trying different values in the equation to see which one fits well. First let’s try x=3.

That is too small; we’re looking for the outcome to be 40. Let’s try x=3.5, we’re told that x=3.5 is the highest limit for which the value of x can be to result in 40.

That is too big we need to try a smaller value. Let’s try x=3.3

The value is too small, the result is very close to 40 which means we must be very close to the value that we’re after. Let’s try x=3.4

x=3.4 provides a very large outcome of 42.704 which is far from what we’re after. How about we try x=3.35.

The above informs us that the solution must be between x=3.3 and x=3.35 but x=3.35 is too big. This must mean that the solution is less than x=3.35 but not below x=3.3
This must mean that the solution is x=3.3 because we’re asked to round to one decimal place.