View Single Post
Old 2004-04-27, 14:42   #21
jinydu's Avatar
Dec 2003
Hopefully Near M48

2·3·293 Posts

I think there was somewhat of a misunderstanding. This is what I meant:

We all know that the solution to the quadratic equaion ax^2 + bx + c = 0 is x=(-b+-sqrt(b^2 - 4ac))/2a. Not only does the quadratic formula give you the root of the polynomial, it also allows you to simplify your answer to the fullest extent possible. That is, in "most" cases, the answer will look something like: x = p+sqrt(r). Now, if x happens to be a rational number, then taking the square root of r will yield a rational number. After just a few simple operations, you can express x in the form a/b, where a and b are integers. In the even simpler case where x is an integer, the numerator will clearly divide the denominator.

For example, with the quadratic formula applied to x^2 - 2x + 1 = 0, x = (-(-2)+-sqrt((2^2)-(4*1*1)))/(2*1). But this can be simplified:

x = (2+-sqrt(4-4))/2
x = (2+-sqrt(0))/2
x = (2+-0)/2
x = 2/2
x = 1

This is what I mean by simplifying it to the simplest form, rather than leaving the answer as x = x^2 - 2x + 1 = 0, x = (-(-2)+-sqrt((2^2)-(4*1*1)))/(2*1).

Originally Posted by jinydu
And in general: The cubic formula tends to give answers that look something like cube root (a+sqrt(b)) + cube root (a-sqrt(b)) + c. In the particular cases where the solution has a simpler form (such as an integer, rational number or expression with only a single radical), how can I get to this simpler form?
Basically, what I'm looking for is an analogous process for the cubic roots (which have already been calculated using the cubic formula). There are some special cases that are trivial to simplify. For example, if a=sqrt(b), then x=cbrt(a+sqrt(b))+c. But there are less trivial cases, such as when x is an integer. I would like to know how to simplify to each of the following forms (when they are possible, and some may be subsets of others):

a (integer)
p/q (rational number)

As an analogue to what I said earlier, I am not satisfied with saying that the solution to x^3 + 6x - 20 = 0 is

x = cbrt(10 + sqrt(108)) + cbrt(10 - sqrt(108)),

just as I would be unsatisfied with saying that the solution to x^2 - 2x + 1 = 0 is

x = (-(-2)+-sqrt((2^2)-(4*1*1)))/(2*1).

As in the quadratic example, I would like to go through a series of well-defined steps to arrive at the solution

x = 2.

Last fiddled with by jinydu on 2004-04-27 at 14:50
jinydu is offline   Reply With Quote