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Ferrari's solution to the quartic


Well-known member
Mar 4, 2013
I hope its not bad to ask this.

wikipedia said:
The depressed quartic can be solved by means of a method discovered by Lodovico Ferrari. The depressed equation may be rewritten (this is easily verified by expanding the square and regrouping all terms in the left-hand side)
wikipedia said:
Then, we add a variable y to factor of the left-hand side. This amounts to add some expression to the left-hand side. We add thus the same expression to the right-hand side. After regrouping the coefficients of the power of u in the right-hand side, this gives the equation
which is equivalent to the original equation, whichever value is given to y.

As the value of y may be arbitrarily chosen, we will choose it in order to get a perfect square in the right-hand side. This implies that thediscriminant in u of this quadratic equation is zero, that is y is a root of the equation
What does that mean?

which may be rewritten
The value of y may thus be obtained from the formulas provided in cubic equation.
When y is a root of equation (4), the right-hand side of equation (3) the square of

I also wonder how solving y which is an arbitary number will solve u...

Can anyone explain the whole process,please???

Then I would be soooo thankful....

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
Re: Ferarri's solution to the quartic

Hey mathmaniac! ;)

What does that mean?
The quadratic equation of the right hand side is of the form $Au^2+Bu+C = 0$.
A perfect square is of the form $A(u-D)^2 = 0$.

This means that the solutions for u in this quadratic equation must be identical.
The solution of the quadratic equation is:
$$u=\frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$$
If it only has one solution, then $B^2 - 4AC$ must be zero.
The expression $B^2 - 4AC$ is called the discriminant.

I also wonder how solving y which is an arbitary number will solve u...
Picking a special number for y, makes it easier to solve for u, since we will get an equation of the form
We can take the square root from both sides, leaving us with a regular quadratic equation.

Afterward, the real solution can be constructed from u and y.