# Where to find?

1. Jul 9, 2007

### edgo

L.S.
I am looking for detailed information on the subject of cubic equations. Problem is that I cannot find anything but the most basic properties of such equations. For example, if given an arbitrary cubic equation with roots x(i),where to find the equation like with roots y(i) = x(i) - x(j)? Another one: Omar Khayyam found a construction /graphical solution for the cubic with only one real root. I suppose that since his work also the graphical solution for cubics with three real roots has been found; where is that published, please help me to that!
For a good understanding: I have the answers to both problems but had to calculate them myself. It took a long time and I expect to have done a monks job. Thanks for your help.

2. Jul 10, 2007

### Gib Z

I have no idea where you would find this sort of knowledge other than in a Polynomial's course. However I think I can answer your questions for you, if you clarify a few things.

"arbitrary cubic equation with roots x(i)" does not mean anything to me. Is x(i) some sort of function? If not, you have only given me One root. What is x(j)? Is it another root of the original cubic? Or some other pro numeral?

Given an arbitrary cubic equation with roots $\alpha, \beta, \gamma$, find another cubic with roots $\alpha + C, \beta + C, \gamma + C$ where C is some constant.

I think that is what you mean, so here it is:

Given $$ax^3+bx^2+cx+d$$ has roots $$\alpha, \beta , \gamma$$ we know by definition that
$$a (\alpha)^3 + b (\alpha)^2 + c (\alpha)x + d = 0$$
and similarly for beta and gamma.

We can see that is we replaced x with alpha, we get 0. So if we wanted to replace x with alpha + C and still get 0, we have to compensate for the extra C that was added, in other words instead of being just x, it would become the new polynomial with the new roots is
$$a (x-C)^3 + b(x-C)^2 + c(x-C) + d$$, which you can expand to simplify.

Lets have an example, the cubic $$x^3 - x$$. It has some root, alpha. If we wanted a cubic with a root alpha - 1, then the new polynomial would be $$(x+1)^3 - (x+1)$$ which simplifies to $$x^3 + 3x^2 + 2x$$.

The original cubic had roots 0, 1 and -1. So this new polynomial should have roots -1, 0 and -2. You can see for yourself that it is true.

As for the second one concerning the graphical solution to cubic, Omar Khayyam for a graphical solution for a cubic which only found 1 real root. That means it could also have been used to find one of the real roots of cubics that have 1 2 or 3 real roots. As long you know one root, you can do some polynomial division and use the quadratic formula for the others.

3. Jul 10, 2007

### Schrodinger's Dog

Last edited: Jul 10, 2007
4. Jul 10, 2007

### edgo

Thanks for the reactions to my question. The Wikipedia links do exactly what I'm trying to avoid: they give the Cardano or Harriot or the goniometric solutions to the cubic polynomials. I'm interested in a "general theory" on the cubic polynomial. It was university stuff till the first decades of the 20th century, so to say before math. exploded into a world of disciplines. I do have some books which were used those days and they have helped me in the beginning. It's a pity that that kind of math. is neglected these days, there is a world of discoveries waiting. Cardano is giving just a solution, it's not the end to it.
By the way, I posted my question in the pre-calculus group only because calculus has nothing to do with it. Maybe not my best start.
I am used to MS Word equation editor and found PF to reject that script (why?). I am sorry that you misunderstood my "emergency notation". I'll try it with LateX:
$$V_{1}=x^3+ax^2+bx+c=0$$ is an arbitrary cubic equation with roots

$$x_{1}$$ , $$x_{2}$$ and $$x_{3}$$.

I was referring to the cubic equation with roots

$$y_{1}$$ = $$x_{1}$$ - $$x_{2}$$ etc. Again, sorry for that.

5. Jul 10, 2007

### Gib Z

When you say "I was referring to the cubic equation with roots" and then only give me one root it confuses me :(

6. Jul 10, 2007

### edgo

It was the root x with the general index i where i = 1, 2 or 3. Also the three roots in a general notation. I think it's not unusual to do so.

7. Jul 10, 2007

### edgo

Sorry, your question was another one.
$$y_{1}=x_{1}-x_{2}$$
$$y_{2}=x_{2}-x_{3}$$
$$y_{3}=x_{3}-x_{1)$$

8. Jul 10, 2007

### Gib Z

For that I am quite certain that you need to know the values of $x_1, x_2, x_3$, otherwise it is not possible.

9. Jul 11, 2007

### edgo

Sorry to disappoint you but you don't have to know $$x_i$$. It's just a matter of coefficients.