MHB Ferrari's solution to the quartic

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The discussion centers on Ferrari's method for solving the depressed quartic equation, emphasizing the manipulation of variables to achieve a perfect square on the right-hand side. By introducing an arbitrary variable y, the equation can be simplified, allowing for the determination of u through a quadratic form. The key point is that selecting a specific value for y leads to a discriminant of zero, resulting in a unique solution for u. This process illustrates how an arbitrary choice can facilitate solving complex equations. Understanding this method is crucial for grasping the intricacies of quartic equations.
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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:
b692bead5cbf59bcdbc0699875161a2d.png
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
549f3239c9359dcd9d2e988cba906203.png
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?

8497b88e1060cbbe181cfeb45583d7f1.png
which may be rewritten
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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
348e62c5af1d54b08560a55d46220f4b.png


........

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...
 
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Re: Ferarri's solution to the quartic

Hey mathmaniac! ;)

mathmaniac said:
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.
 
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