The first non-linear equation encountered in algebra courses is usually the quadratic equation
and all students will be familiar with the formula for its roots:
The formula for the roots of a general cubic is somewhat more complicated and that for a general quartic usually takes several pages to describe! We are spared further effort by a theorem which states that there is no such formula for general polynomials of degree higher than four. Accordingly, except in special cases (for example, when factorization is easy), we prefer in practice to use a numerical method to solve polynomial equations of degree higher than two.
Another class of nonlinear equations consists
of those which involve transcendental functions such as 
. Useful analytic solutions of such equations are rare so that we are usually forced to use numerical methods.
We shall use a simple mathematical problem to show that transcendental equations do arise quite naturally. Consider the height of a liquid in a cylindrical tank of radius r and horizontal axis, when the tank is a quarter full (see Figure 2). Denote the height of the liquid by h (DB in the diagram). The condition to be satisfied is that the area of the segment ABC should be ¼ of the area of the circle. This task reduces to
,
where 
is the area of the sector OAB, the
triangle OAD. Hence
.
When we have solved the transcendental equation
we obtain h from
FIGURE
2.
Cylindrical tank (cross-section).
Let it be required to find some or all of the roots of the nonlinear f(x) = 0. Before we use a numerical method (STEP 7, STEP 8, STEP 9 and STEP 10), we should have some idea about the number, nature and approximate location of the roots. The usual approach involves the construction of graphs and perhaps a table of values of the function f, in order to confirm the information obtained from the graph.
We will now illustrate this approach by a few examples.
a) 
If we do not have a calculator or computer available to immediately plot the graph of
f(x )= sin x - x + 0.5,
we
can separate f into two parts, sketch two curves
on a single set of axes, and find out whether they
intersect. Thus we sketch 
.
Since 
, we are only interested in the
interval -0.5 £ x £ 1.5 (outside which |x - 0.5| > 1).
Thus we deduce from Fig. 3 that the equation has only one
real root, near x =1.5 as follows:
We now know that the root lies between 1.49 and 1.50, and
we can use a numerical method to obtain a more accurate
answer as is discussed in tlater Steps.
b) 
Again, we sketch two curves:
In order to sketch the second curve, we
use the three obvious zeros at x = 0, 2, and 3, as well
as the knowledge that x(x - 2) (x - 3) is negative
for x < 0 and 2 < x < 3, but
positive and increasing steadily for x > 3. We
deduce from the graph (Fig. 4) that there are three real
roots, near x = 0.2, 1.8, and 3. 1, and tabulate
as follows (with 
:
We conclude that the roots lie between 0.15 and 0.2, 1.6 and 1.8, and 3.1 and 3.2, respectively. Note that the values in the table were calculated to an accuracy of at least 5SD. For example, working to 5S accuracy, we have f (0.15) = 0.97045- 0.79088= 0.17957, which is then rounded to 0.1796. Thus the entry in the table for f(0.15) is 0.1796 and not 0.1795 as one might expect from calculating 0.9704 - 0.7909.
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