Knowledge of the quadratic formula is older than the Pythagorean
Theorem. Solving a cubic equation, on the other hand, was the first major
success story of Renaissance mathematics in Italy.
The solution was first published by
Girolamo Cardano (1501-1576) in his Algebra book Ars Magna.
Our objective is to find a real root of the cubic equation
The other two roots (real or complex) can then be found by polynomial division and the quadratic formula. The solution proceeds in two steps. First, the cubic equation is "depressed"; then one solves the depressed cubic.
This trick, which transforms the general cubic equation into a new cubic equation
with missing x2-term is due to
Nicoḷ Fontana Tartaglia (1500-1557). We apply the substitution
to the cubic equation, to obtain:
Multiplying out and simplifying, we obtain the "depressed" cubic
Let's try this for the example
Our substitution will be x=y+5; expanding and simplifying, we obtain the depressed cubic equation
We are left with solving a depressed cubic equation of the form
How to do this had been discovered earlier by
Scipione del Ferro (1465-1526).
We will find s and t so that
It turns out that y=s-t will be a solution of the depressed cubic. Let's check that:
Replacing A, B and y as indicated transforms our equation into
This is true since we can simplify the left side by using the binomial formula to:
How can we find s and t satisfying (1) and (2)? Solving the first equation for s and substituting into (2) yields:
Simplifying, this turns into the "tri-quadratic" equation
which using the substitution u=t3 becomes the quadratic equation
From this, we can find a value for u by the quadratic formula, then obtain t, afterwards s and we're done.
Let's do the computation for our example
We need s and t to satisfy
Solving for s in (3) and substituting the result into (4) yields:
which multiplied by t3 becomes
Using the quadratic formula, we obtain that
We will discard the negative root, then take the cube root to obtain t:
By Equation (4),
Our solution y for the depressed cubic equation is the difference of s and t:
The solution to our original cubic equation
is given by
I will not discuss a slight problem you can encounter, if you follow the route outlined. What problem am I talking about?
Shortly after the discovery of a method to solve the cubic equation,
Lodovico Ferrari (1522-1565), a student of Cardano, found a similar method to solve the quartic equation.
This section is loosely based on a chapter in the book Journey Through Genius
by William Dunham.
Show that y=2 is a solution of our depressed cubic
Then find the other two roots. Which of the roots equals our solution
Transform the cubic equation
into a depressed cubic.
Find a real root of the cubic equation in Exercise 2.
(This is for practice purposes only; to make the computations a little less messy, the root will turn out to be an integer, so one could use the Rational Zero test instead.)
[Next: The Geometry of the Cubic Formula]
S.O.S MATHematics home page
Do you need more help? Please post your question on our
S.O.S. Mathematics CyberBoard.
Copyright © 1999-2017 MathMedics, LLC. All rights reserved.
Math Medics, LLC. - P.O. Box 12395 - El Paso TX 79913 - USA
users online during the last hour