Solving Equations
This page shows you how to solve equations using trial and estimation and the Iteration method.
Trial and Improvement
Any equation can be solved by trial and improvement (/error). However, this is a tedious procedure. Start by estimating the solution (you may be given this estimate). Then substitute this into the equation to determine whether your estimate is too high or too low. Refine your estimate and repeat the process.
Example
Solve t³ + t = 17 by trial and improvement.
Firstly, select a value of t to try in the equation. I have selected t = 2. Put this value into the equation. We are trying to get the answer of 17.
If t = 2, then t³ + t = 2³ + 2 = 10 . This is lower than 17, so we try a higher value for t.
If t = 2.5, t³ + t = 18.125 (too high)
If t = 2.4, t³ + t = 16.224 (too low)
If t = 2.45, t³ + t = 17.156 (too high)
If t = 2.44, t³ + t = 16.966 (too low)
If t = 2.445, t³ + t = 17.061 (too high)
So we know that t is between 2.44 and 2.445. So to 2 decimal places, t = 2.44.
Iteration
This is a way of solving equations. It involves rearranging the equation you are trying to solve to give an iteration formula. This is then used repeatedly (using an estimate to start with) to get closer and closer to the answer.
An iteration formula might look like the following (this is for the equation x^{2} = 2x + 1):
x_{n+1} | = | 2 | + |
1 |
x_{n} |
You are usually given a starting value, which is called _{0}. If x_{0} = 3, substitute 3 into the original equation where it says x_{n}. This will give you x_{1}. (This is because if n = 0, x_{1} = 2 + 1/x_{0} and x_{0} = 3).
x_{1} = 2 + 1/3 = 2.333 333 (by substituting in 3).
To find x_{2}, substitute the value you found for x_{1}.
x_{2} = 2 + 1/(2.333 333) = 2.428 571
Repeat this until you get an answer to a suitable degree of accuracy. This may be about the 5th value for an answer correct to 3s.f. In this example, x_{5} = 2.414...
Example
a) Show that x = 1 + 11
x - 3
is a rearrangement of the equation x² - 4x - 8 = 0.
b) Use the iterative formula:
x_{n+1} | = | 1 | + |
11 |
x_{n} - 3 |
together with a starting value of x_{1} = -2 to obtain a root of the equation x² - 4x - 8 = 0 accurate to one decimal place.
a) multiply everything by (x - 3):
x(x - 3) = 1(x - 3) + 11
so x² - 3x = x + 8
so x² - 4x - 8 = 0
b) x_{1} = -2
x_{2} = 1 + 11 (substitute -2 into the iteration formula)
-2 - 3
= -1.2
x_{3} = 1 + 11 (substitute -1.2 into the above formula)
-1.2 - 3
= -1.619
x_{4} = -1.381
x_{5} = -1.511
x_{6} = -1.439
x_{7} = -1.478
therefore, to one decimal place, x = 1.5 .