Solving general quartic equations

A new method of solving general quartic equations building on quasi-symmetric equations is described.

[pseudoexpertise]

This video is indeed original. I've once (about a year ago) been brute-forcing through all the math channels I could possibly find and got about 600. Then I went through them once by each and purged out all the trashy/too elementary ones. With that only 50 channels were left.
What I can ensure: There has not been a single video of that kind! I can confirm this video's originality!
Great video!

[cblpu]

Excellent video. i wish videos like these got more recognition

[Tntss-gv]

Great, video and very smart method. As I myself worked on quartics I didn't think about this way of solving, good work.
As a Ferrari's method user, I find his method really good because you don't have to memorize an entire equation, just the appropriate substitutions. So the only drawback I could find about your method is of course the entire cubic equation to memorize. But it's still really good !

[finnboltz]

This is called the 'Christianson-Brown method' for finding the solution to a quartic. It was discovered in 1993.

[detectiveandspynovels]

Sir Marvel. You are Captain Marvel of Maths, your are Thor of Maths, you are Martian Man Hunter of Math. Sir I have checked each and every equation.Each time I say, Sir you are great

[Farshaadaa]

In f(X) = X^4+aX^3+bX^2+cX+d =0 we could check the coefficient and if 1+ a+ b+ c+ d=0, X=1 that is one answer, then dividing f(X) by X-1 getting a Cubic equation to solve.. Likewise if 1- a+ b- c+ d=0 we have X= -1 that is one answer, and as it mentioned above, a Cubic equation needs to be solved for the other answers.

When studying elliptic curves which can be represented by y^2 = x^4 + ax^3 + bx^2 + cx + d, for example, this approach follows naturally. Good work finding this if you haven't studied elliptic curves. Good even if you have.

[MutangamweDjsgzbsvcsf-gocr]

Your reasonning great but I think the polynomial equations like x^4+x^3+2x^2+3x+4=0 may not work. If yes, how would I manipulate it to get its solutions.

[XJWill]

This method has some similarities to the Ferrari method, but it is considerably simpler to describe and use. Consider the depressed quartic equation:
y^4 + u*y^2 + v*y + w = 0
If v = 0, then the equation is quadratic in y^2 and easily solved. So assume v is non-zero for this method.
Rearrange to
y^4 = -u*y^2 - v*y - w

[hamidkh]

Thank you.
I prefer this way to Ferrari's method. I think this way, has less calculations.
a good start . keep going

[simeonakinleye]

Is it that it has no root or no real roots, when the curve touches or cross the real line?

[holyshit]

I tried to reduce general quartic to the biquadratic case and got equation of sixth degree
Suppose we have equation + a_{0} = 0
Lets substitute x = (pt+q)/(t+1) and multiply equation by (t+1)^4
You will get quartic in t
Equate coefficients in terms t^3 and t to zero
and you will get system of equation
Solution to this system of equations leads us to polynomial equation of 10th degree
but this 10th degree polynomial is divisible by polynomial + a_{0}
so there is polynomial equation of sixth degree left to solve

[XJWill]

For the video, I found the animation to be distracting. I would have rather just seen each new line appear instantly, and then a pause for viewing, and then the next new line appear instantly.
That is an interesting method to solve general quartics. I have not seen it before. All methods I have seen to solve general quartic equations involve solving a cubic equation. So when comparing quartic solution methods, it is essentially a question of how complicated or tedious it is to determine the cubic equation and then apply the root of the cubic to solve the quartic. I think the method in the video compares favorably with the Ferrari method, but perhaps is slightly behind the method I will outline below in another comment.

[giuseppeimbimbo]

I tried to resolve the equation
x^4-7x^2+11x-10=0
One of the real roots is x=2
The equation has another real root
Can we find the exact value by using this brand new method?