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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.

2. 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.

3. 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.


4. 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.