# Math Is Fun Forum

Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫  π  -¹ ² ³ °

You are not logged in.

## #1 2012-04-19 19:55:34

anonimnystefy
Real Member
From: Harlan's World
Registered: 2011-05-23
Posts: 16,049

### Solving cubic equations in general

The method I am describing today was given by Girolamo Cardano-an Italian mathematician,who first showed this method in his book Ars Magna.

Let us attempt to solve the general cubic equation:

Later,I will show you how this method is used for some cubic equations.

The first step of solving the equation is called "Depressing the cubic equation".Its purpose is to use a handy substitution to bring the equation down to a much simpler form.

The substitution we are using is

which gives us the equation:

When we multiply everything out we get the equation:

which doesn't look any easier or healthier to solve than the starting equation,but if you look a little closer you can see that the equation is of the form:

which we obtain by moving the free term to the RHS.

The next method was discovered by a mathematician called Scipione dal Ferro.

He said that the equation will be solved if we find such numbers s and t so that the next two equations hold:

He also said that one solution to the cubic (3) is s-t.You can check this by plugging in the values:

But this is still a problem.We don't know how to solve the system for s and t.How are we closer to the answer? Well,we can just express s in terms of t from the equation (4) and just plug it in into the equation (5).

Let's see what we get:

Expanding the equation we obtain a nice "triquadratic" equation, analogous to the biquadratic equation of the fourth degree.Here it is:

Using the substitution

we get:

We can now back substitute this into (4) to get s.From this we can get y=s-t.Then we back substitute y into (2) from which we get x.We now have at least one solution to the starting cubic equation.We can get other roots by polynomial division and quadratic formula.

Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.

Offline

## #2 2012-04-19 20:10:32

Bob
Registered: 2010-06-20
Posts: 10,544

### Re: Solving cubic equations in general

hi Stefy,

Thanks for posting this.

Very clear explanation.

Bob

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Offline

## #3 2012-04-19 20:20:16

anonimnystefy
Real Member
From: Harlan's World
Registered: 2011-05-23
Posts: 16,049

### Re: Solving cubic equations in general

Thank you,Bob!

For a little detailed explanation,you can look at this page:Method of Solving General Cubic Equations

Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.

Offline

## #4 2012-08-13 00:33:34

anonimnystefy
Real Member
From: Harlan's World
Registered: 2011-05-23
Posts: 16,049

### Re: Solving cubic equations in general

Here are some example equations (some of them are simpler if you do not use this method, but they are a nice practice):

Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.

Offline

## #5 2014-02-02 20:23:38

Yusuke00
Member
Registered: 2013-11-19
Posts: 43

### Re: Solving cubic equations in general

Also you could use Viette's formulas and the fact that f=(X-x1)(X-x2)(X-x3) where x1,x2,x3 are roots of the ecuation.

Offline

## #6 2014-02-03 01:11:36

Nehushtan
Member
Registered: 2013-03-09
Posts: 957

### Re: Solving cubic equations in general

Yusuke00 wrote:

Also you could use Viette's formulas and the fact that f=(X-x1)(X-x2)(X-x3) where x1,x2,x3 are roots of the ecuation.