Question

How to solve the quadratic equation with the Tschirnhaus transformation.

Answer 1

The Tschirnhaus transformation is a transformation commonly used in cubics and quartics (and polynomials of higher degree) to reduce the coefficient of the term of the second highest degree to zero. In other words, when applied on the quadratic, it gets rid of the "bx" term.

Answer 2

Assuming one has an equation ax^2+bx+c=0, the first step would be to divide by a, because the a is an unneeded constant. For brevity's sake, I will rewrite the quadratic in the following form
\[x^2+ax+b=0\]
Now, we shall apply the Tschirnhaus transformation.

Answer 3

The Tschirnhaus transformation is simply a substitution of x with another value. However, it is a very useful and powerful substitution.
We substitute x by using the equation \[x=u-(1/2)a\] (The reasoning behind this will be later explained)
u in this case is a variable defined by u=x+(1/2)a, a being the coefficient of the first degree term. We shall now see what this results in.

Answer 4

\[(u-(1/2)a)^2+a(u-(1/2)a)+b=0 \]
\[u^2-au+(1/4)a^2+au-(1/2)a^2+b=0\]
The "magic" behind this substitution is that we get rid of the first degree u term! Because a is a constant (the coefficient of the term of the first degree, ax), we have a quadratic equation where we can just move everything to one side and solve!
\[u^2-(1/4)a^2+b=0\]
\[u^2=(1/4)a^2-b\]
\[u=\pm \sqrt{(1/4)a^2-b}\]

Answer 5

We have just solved this equation for u.
All that is necessary is to resubstitute u.
\[u=x+(1/2)a\]
\[x+(1/2)a=\pm \sqrt{(1/4)a^2-b}\]
\[x=-(1/2)a\pm \sqrt{(1/4)a^2-b}\]
And that is the solution to the quadratic equation in the form x^2+bx+c=0.

Answer 6

Now, comes the big question: Why substitute \[u=x+(1/2)a\]?
Let us apply the Tschirnhaus transformation to a polynomial of degree n. Remember, the Tschirnhaus transformation is used to reduce the term with the second highest degree, so terms close to constants are irrelevant.
\[x^n+ax^{n-1}+bx^{n-2}+cx^{n-3}+...=0\]
The Tschirnhaus transformation is the substitution of
\[x=u-(1/n)a\]
Let us attempt this on this generalized polynomial of nth degree.
\[(u-(1/n)a)^n+a(u-(1/n)a)^{n-1}+b(u-(1/n)a)^{n-2}+c(u-(1/n)a)^{n-3}+...=0\]
This yields (we focus only on the first few terms of the expansion)
\[[u^n-au^{n-1}...]+[a[u^{n-1}-...]]+b(u-(1/n)a)^{n-2}+c(u-(1/n)a)^{n-3}+...=0\]
In fact, the latter terms are irrelevant. We shall soon see why.

Answer 7

this is fascinating lol *_* when do they teach these stuffs

Answer 8

\[[u^n-au^{n-1}+[au^{n-1}...]\] Right here.
\[au^{n-1}\]
Cancels out brilliantly. The result? We effectively get rid of the term that is one less than the degree of the polynomial.

Answer 9

Now, one would probably ask at this point, if we could apply the Tschirnhaus transformation to any polynomial, could we solve any polynomial? The answer to this question is not a yes or no. In fact, one must notice that the Tschirnhaus transformation does not reduce all the terms of a polynomial, only one. It is not really possible to apply another transformation to reduce the polynomial further.
There are, in fact, solutions to the cubic as well as the quartic, in the sense that they are expressed in finite sums, differences, products, and nth-roots (n being a natural number). In fact, the Tschirnhaus transformation is the first step towards solving them.
What about the quintics, and the sextics, and polynomials of higher degree? The answer is no, in the sense of how one would solve the cubic and quartic.

Answer 10

In fact, the proof that there is no closed form general solution for all cases of polynomials of degree 5 or higher is an application of Group Theory, a fairly advanced mathematical subject.

Answer 11

160 minutes :/ But, I'm about to go to bed :D

Answer 12

@inkyvoyd Wow:D Great tutorial. I learnt a new transformation today:D

Answer 13

Yea, it took me forever to understand this, because Wikipedia barely goes through it.

Answer 14

Actually read a (translated?) version of Euler's elements of algebra, and it showed the solution to the cubic and quartic, although I read only the part about the cubic.

Answer 15

Wow:) You have done a great work. Very informative I must say:)

Answer 16

I'm going to try to show how to solve the cubic tomorrow, but I'll have to do a bit more clarification myself :)
I'm going to use Cardano's method (which needs this substitution), but there are a lot others, just many are quite hard.

Answer 17

*more like a few, but I don't really understand them myself, so ...

Answer 18

I'm basically messaging everyone I know that is "good" at math, because I'm curious as to why many people haven't heard of this transformation

Answer 19

very nice! ,i used this substitution without knowing in a problem, (reducing a cubic), i did not realise all this though :D

Answer 20

You did? So you do know about it!

Answer 21

When did you learn about solving the cubic? and, if you did, the quartic?

Answer 22

last year my teacher challenged me to solve the cubic :S i made some progress, i had this substitution and i used some kind of identity, which got me somewhere. i'll try and remember it

Answer 23

\[x^3 + ax^2 + bx + c = 0\]

let \[x = u - \frac{a}{3}\]

Answer 24

i think

Answer 25

took me so long to get anywhere with it though

Answer 26

as instead of making a general substitution u + t and soliving to eliminate a

i just kept trying different ones

Answer 27

So, do you know how to solve the cubic? @eigenschmeigen

Answer 28

320 mins till I can bump again, so hopefully I get people to read about it.

Answer 29

For further reading, there is this paper that I found on my favorites: http://www.sigsam.org/bulletin/articles/145/Adamchik.pdf
And there's an entry at the Proof Wiki also :-)
Great work @inkyvoyd

Answer 30

Time for the cubic :S

Answer 31

An application of this transformation is Cardano's solution the the cubic equation. I have also explained how to solve it at
http://openstudy.com/study#/updates/4fa79c8be4b029e9dc387b14 .