18 1. Fundamentals
(b) Determine by inspection a root of the polynomial equation
t3 - 4t + 3 = 0,
and use this information to find a complete set of solutions to
the equation.
- Consider the cubic equation
x3 - 12x2 +29x - 18 = 0.
As we shall see below, it is possible to solve cubic equations in which
the quadratic coefficient vanishes. Fortunately, a simple transforma-
tion permits us to reduce any cubic to this form. Verify that the
substitution x = t + 4 converts the equation to
t3 - 19t - 30 = 0.
By inspection, obtain a solution to the equation in t, and thence solve
the equation in x.
- The solutions of the cubic equations so far have involved finding one
solution by guessing. This is too much to expect in general. Argue
that, if a general method can be found to solve cubic equations of the
form
ts+pt+q=o,
then it is possible to solve any cubic equation. - The cubic equation: Cardan’s Method. An elegant way to solve the
general cubic is due to Cardan. The strategy is to replace an equation
in one variable by one in two variables. This provides an extra degree
of freedom by which we can impose a convenient second constraint,
allowing us to reduce the problem to that of solving a quadratic.
(a) Suppose the given equation is
t3 + pt + q = 0.
Set t = u + v and obtain the equation
u3+v%-(3uv+p)(21+v)+q=O.
Impose the second condition 3uv + p = 0 (why do we do this?)
and argue that we can obtain solutions for the cubic by solving
the system
u3+v3= -q
uv = -p/3.