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Ferrari's solution to the quartic

mathmaniac

Active member
Mar 4, 2013
188
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
8,776
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
$$(u^2+\alpha+y)^2=A(u-D)^2$$
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.