See Also: Quadratic Equations and
Simultaneous Equations
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):
You are usually given a starting value, which is called x_{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 .
Copyright © Matthew Pinkney 2003
