Hatrack River Forum: Quadratic Equasions and More...

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Posted by Jonathan Howard (Member # 6934) on :

We've studied quadratic equasions at school:

code:

ax^2 + bx + c = 0 /4a
4a^2x^2 + 4abx + 4ac = 0 /- 4ac + b^2
4a^2x^a + 4abx + b^2 = b^2 - 4ac
(2ax + b)^2 = b^2 - 4ac /SQRT()
2ax + b = +/- SQRT(b^2 - 4ac) /- b
2ax = - b +/- SQRT(b^2 - 4ac) /: 2a
x = [-b +/- SQRT(b^2 - 4ac)]:2a

But, how does one solve a "cubatic" equasion? I tried simplifying:

code:

ax^3 + bx^2 + cx + d = 0
x(ax^2 + bx + c) + d = 0

But... If I get rid of the d, it does not help me, because after dividing by x, it's d:x on the other side. As for dividing by D and getting rid of the "1", I can't say that when x(ax^2 + bx + c) = -1 either x or ax^2 + bx + c equals to -1, because it only works like that with zero or infinity.

Ergo, how do I solve a "cubatic" equasion?

JH

[ February 06, 2005, 10:42 AM: Message edited by: Jonathan Howard ]

Posted by Dagonee (Member # 5818) on :

A polynomial equation may be solved either numerically (trying different values until an intercept is found - there's a method for selecting the next guess that makes this less difficult than it sounds) or by factoring.

The first method is really only suitable for computers. My first assignment in numerical methods was to write a program to solve n-order polynomial equations to 200 significant digits (which meant we had to write a math package from scratch).

There are formulas for polynomial equations of orders 1-4:

quote:
Polynomials of orders one to four are solvable using only rational operations and finite root extractions. A first-order equation is trivially solvable. A second-order equation is soluble using the quadratic equation. A third-order equation is solvable using the cubic equation. A fourth-order equation is solvable using the quartic equation. It was proved by Abel Eric Weisstein's World of Biography and Galois Eric Weisstein's World of Biography using group theory that general equations of fifth and higher order cannot be solved rationally with finite root extractions (Abel's impossibility theorem).

However, the general quintic equation may be given in terms of the Jacobi theta functions, or hypergeometric functions in one variable. Hermite Eric Weisstein's World of Biography and Kronecker Eric Weisstein's World of Biography proved that higher order polynomials are not soluble in the same manner. Klein showed that the work of Hermite was implicit in the group properties of the icosahedron. Klein's method of solving the quintic in terms of hypergeometric functions in one variable can be extended to the sextic, but for higher order polynomials, either hypergeometric functions in several variables or "Siegel functions" must be used (Belardinelli 1960, King 1996, Chow 1999). In the 1880s, Poincaré Eric Weisstein's World of Biography created functions which give the solution to the nth order polynomial equation in finite form. These functions turned out to be "natural" generalizations of the elliptic functions.

See http://mathworld.wolfram.com/Polynomial.html

Posted by Icarus (Member # 3162) on :

quote:
A polynomial equation may be solved either numerically (trying different values until an intercept is found - there's a method for selecting the next guess that makes this less difficult than it sounds) or by factoring.

The first method is really only suitable for computers. . . .

There's kind of a combination method that is not so impossible. For a polynomial equation y = Ax^n + . . . . + B, you can, first of all, use desCartes's Rule of Signs to get a sense of how many positive and how many negative roots there may be. Then, you can make a list of all the possible rational roots, because all the possible rational roots must be of the form

code:

 
        a factor of B
root = ---------------
        a factor of A

(I'm too lazy to explain why now, but if you can't figure it out and are itching to know, I'll be happy to do so.)

So, between these two tidbits, you have a finite list of possibilities to try. You try these numbers out, not by plugging them into the polynomial and seeing if you get a zero, but by dividing the polynomial by (x – #). [And you can most comfortably do this using synthetic division.] When you get a remainder of zero, you have not only found a root, but you have also simplified the remaining polynomial.

If the degree of the polynomial was n, then if you can find at least n – 2 roots in this manner, you can find them all, because the final quotient will be quadratic, and if you can't find the roots by factoring, you can use the quadratic formula. If you can't get it down to the final two roots (and you don't know any of the other formulae, which are commonly not taught at your level), then you have a lot of irrational roots, and you are left with educated guess-and-check (again, there are methods to simplify your life here) to approximate irrational roots, and this can be tedious work even under the best of circumstances.

Posted by Icarus (Member # 3162) on :

by the way, it's "equations"

Also, I think it's really cool that you're exploring these issues beyond what you have learned in class.

Posted by Jonathan Howard (Member # 6934) on :

Soz, Ick. I get confused between tion ans sion.

Thanks!

JH

Posted by Dagonee (Member # 5818) on :

From a programmer's standpoint, it's easier and less computationally intensive to use the Newton's method, which is the method we used. Certainly, it's not a "method" to simply start at an arbitrary number and keep guessing until you get to zero.

Edit: Didn't mean to denigrate your explanation - If I had to do it by hand, that's what I'd do. I'd totally forgotten about it.

Dagonee

[ February 06, 2005, 11:35 AM: Message edited by: Dagonee ]

Posted by King of Men (Member # 6684) on :

Why not Google for 'cubic equation'? That should give you some solution methods.

Posted by Dagonee (Member # 5818) on :

If he uses my link he'll find the formula.

Posted by HollowEarth (Member # 2586) on :

While its possible to solve a general polynomial algebraically up to the 4th degree, above the second or maybe third (in rare cases) its just not worth it by hand. The expressions for the roots are kinda big.

Posted by MidnightBlue (Member # 6146) on :

1. Buy a TI-89.
2. Enter solve(*equation goes here*, x)
3. Hit enter.
4. Read solution(s).

Posted by HollowEarth (Member # 2586) on :

Oh and my understanding is that generally speaking computers use bisection to home in on the root and then use newtons's method. This helps get rid of problem points since we can trick newton's method.

Consider:

code:

f(x) = 1 - x + 0.5 * x ^ 3

f'(x) = -1 + 3/2 * x ^ 2

newtons's method: xn+1 = xn - f(xn) / f'(xn)

We'll use 0 as our guess.

code:

x0 = 0;
x1 = 1;
x2 = 0;
x3 = 1;
x4 = 0;
...

By using bisection first we can get by most of these kind of problems.

Posted by Dagonee (Member # 5818) on :

Yep, that's true. Our algorithm did all that. Of course, it also watched for non-convergence and would try a new starting point if that happened. I forget exactly how we took it into account, but we ran a couple thousand randomly generated polynomial equations through it to test it.