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

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,606

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:

which is a quadratic equation easily solvable by the quadratic formula.

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

Offline

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,469

hi Stefy,

Thanks for posting this.

Very clear explanation.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,606

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

Offline

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,606

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

Offline

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

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

Pages: **1**