See Also: Quadratic Equations and
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
Solve t³ + t = 17 by trial and improvement.
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
If t = 2.5, t³ + t = 18.125 (too high)
If t = 2.4, t³ + t = 16.224 (too
If t = 2.45, t³ + t = 17.156 (too high)
If t = 2.44, t³ + t = 16.966
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.
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.
iteration formula might look like the following (this is for the equation
x2 = 2x + 1):
You are usually given a starting value, which is called x0. If x0 = 3, substitute 3 into the
original equation where it says xn. This will give you
x1. (This is because if n = 0, x1 = 2 + 1/x0 and x0 = 3).
x1 = 2 + 1/3 = 2.333 333 (by
substituting in 3).
To find x2, substitute the value you
found for x1.
x2 = 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, x5 =
a) Show that x = 1 +
x - 3
is a rearrangement of the
equation x² - 4x - 8 = 0.
b) Use the iterative formula:
together with a starting value of x1 = -2 to obtain a root of the
equation x² - 4x - 8 = 0 accurate to one decimal place.
everything by (x - 3):
x(x - 3) = 1(x - 3) + 11
so x² - 3x = x + 8
x² - 4x - 8 = 0
b) x1 = -2
x2 = 1 + 11
(substitute -2 into the
-2 - 3
x3 = 1 + 11
(substitute -1.2 into the above formula)
-1.2 - 3
x4 = -1.381
x5 = -1.511
x6 = -1.439
x7 = -1.478
therefore, to one
decimal place, x = 1.5 .
Copyright © Matthew Pinkney 2003