- #1

**[SOLVED] 3rd degree polynom**

if given Axx+Bx+C=0 then x=f(A,B,C)=(-B(+/-)sqrt(BB-4AC))/2A.

Now given:Axxx+Bxx+Cx+D=0

Solve:x=f(A,B,C,D)

MATHCAD PRO 2K solves it but I don't know how.

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- Thread starter ddr
- Start date

- #1

if given Axx+Bx+C=0 then x=f(A,B,C)=(-B(+/-)sqrt(BB-4AC))/2A.

Now given:Axxx+Bxx+Cx+D=0

Solve:x=f(A,B,C,D)

MATHCAD PRO 2K solves it but I don't know how.

- #2

Hurkyl

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The exact solution is not very pretty... do you know complex numbers?

- #3

lethe

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start with

and then make the subsitution

the quadratic term will cancel out, and divide through by

where

next, make the substitution

which is a quadratic in

take the three cube roots of each of those two solutions, giving you 6 solutions, only three of which should be independent. stick each solution into first

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- #4

I wonder if this method can be generalized on whatever-degree polynom?

- #5

StephenPrivitera

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- #6

marcus

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Originally posted by StephenPrivitera

Maybe Lethe won't mind if I steal his thunder and butt into the conversation and say how I think he could have figured that part out. Suppose Lethe had never heard of doing this before and was inventing the method. Or suppose you Stephen were inventing it, and you wanted to remove the quadratic term by a substitution

x = y - K

what K would you use? It is not so hard to find out what K to use!

In the new polynomial the only stuff quadratic in y will be

the term you know about already---namely by

a(y - Z)

and from the cubic term you are going to get

-a3y

so there will be cancelation if b = 3aZ (which decides what Z to use)

---------------------

Footnote you know Pascal's triangle and the old 121 and 1331 and 14641 and so on, so you expand

(y - Z)

and that tells you the quadratic piece of a(y - Z)

If Lethe or you were inventing this (but maybe the persian algebraist/astronomer/poet Omar Khayyam did it in year 1100 or somebody out of 1001 Nights, who knows, Ali Baba or Sheherezade?) getting rid of the quadratic term would be routine. the clever part would be to see that a cubic which is missing a quadratic term can be solved with the quadratic formula. my respect is reserved for the person who saw how to continue and do that.

- #7

marcus

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Wow! Britannica says that indeed Omar Khayyam did write extensively about solving cubic polynomials around 1100 (he also wrote the Rubaiyat) but that he did not use radicals (the square root formula and that).

According to Britannica, the first solution of a cubic by radicals was in Bologna by Scipione del Ferro in 1515.

He passed the method on to his pupils and for a while it was a close kept secret (they did not immediately run to CNN with their discoveries during the Renaissance) and then someone named Tartaglia told someone named Cardano and Cardano published it, in 1545.

So history tells us that getting rid of the quadratic term in a cubic polynomical is the easy part. The hard part, which is solving

the cubic once that has been removed, took this planet, with its peculiar species of evolved two-legged fish, until 1515.

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- #8

Lonewolf

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I wonder if this method can be generalized on whatever-degree polynom?

Unforunately, the only other degree polynomaial we can do this for is of degree 4. This is a consequence of the so-called

- #9

lethe

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Originally posted by StephenPrivitera

like marcus says, that is the easy part. b/3a you can guess by analogy with the quadratic equation: to complete the square of the quadratic equation (and thus solve), you substitute x=y-b/2a. when you move to the cubic equation, simply replace the 2 with a 3. and by the way, this will work again when we go to the quartic equation.

by the way, i didn t figure this formula out. i learned it in high school. but the whole business is more rewarding when you take a course in galois theory, where the most general results are proved.

- #10

This is all right but I think that the last substitution has to be:Originally posted by lethe

start withax^{3}+bx^{2}+cx+d=0

and then make the subsitutionx=y-b/3a

the quadratic term will cancel out, and divide through byaleaving you with

y^{3}+py+q=0

wherep=(3ac-b^{2})/3aandq=(2b^{3}-9abc)/27a^{2}

next, make the substitutiony=z-p/3z, and multiply through byz^{3}:

z^{6}+qz^{3}-p^{3}/27=0

which is a quadratic inz^{3}. solve it using the quadratic formula, which you listed above:

z^{3}=(-q±√(q^{2}+4p^{3}/27))/2

take the three cube roots of each of those two solutions, giving you 6 solutions, only three of which should be independent. stick each solution into firsty=z-p/3zand thenx=y-b/3ato get three solutions for x.

y=z-p/(3az).I checked it with MATHCAD PRO 2K and that's what I got.

- #11

StephenPrivitera

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- #12

lethe

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Originally posted by ddr

This is all right but I think that the last substitution has to be:

y=z-p/(3az).I checked it with MATHCAD PRO 2K and that's what I got.

yes, you are correct, i forgot to divide out by

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