The Renaissance Solution of the Cubic EquationLet us begin with the equation of Del Ferro
Now suppose that we can find number a and b such that
then The important fact to observe here is that x = (a - b) is a solution to this equation, since
and cancelling like terms on the right hand side of the equation we get
The only question we must answer to consider this a complete solution is whether the values a and b can be obtained. But the equations defining a and b are non-linear equations. This is a non-linear system of equations. This non-linear system of equations must be solved to obtain a solution. How do you solve a system of two equations and two unknowns? Substitution? Let us solve for b in terms of a using the equation m= 3ab.
Plug this into the other equation to get
Then
and resembles a quadratic equation. If we let z = a3, then we get a quadratic equation as follows
with solution
or equivalently
It is easy to see that a is the cube root of z, or
If we take
then we get that a3 - b3 = n. Therefore,
Let's use this formula to try to solve
Now n =56 and (n/2)= 28. On the other hand, m=24 and (m/3) = 8. Then (n/2)2 is 784 and (m/3)3 is 512 and so we have
On ther other hand,
and a - b = 641/3 - 81/3 = 2, which is a solution:
Observe that
We can now solve for all the roots using the quadratic formula and we see that x = -1 + (1 - 28)1/2, and x = -1 - (1 -28)1/2 are not real roots.
Homework Assignment 3Solve the equation using the formula to show that 5 is a root. (hint a3 is 27)
To continue see The Renaissance Solution II |