Even though it is uncommon to see questions asking for an analytical solution to equations of degree 3 or 4, they have been asked on the forum. It's also good to know how, in any case.
For the cubic equation, I would discourage simply memorising the general formula; instead, try to understand the method and remember the form of the required substitutions to reduce the cubic to a quadratic.
IMHO, the cubic is still doable by hand. The quartic is often too tedious to contemplate solving analytically (without a computer), but the method is instructive.
Hootenanny
Apr30-06, 01:02 PM
I hate solving cubic equations, I usually use an itritive method:blushing: Good tutorials though :biggrin:
~H
Hurkyl
Apr30-06, 03:24 PM
I guess it's nitpicky, but I don't think you meant "analytical": you meant solutions in terms of radicals.
There are more general analytic techniques: for example, an explicit (albeit extraordinarily long) solution to the general quintic can be written in terms of hypergeometric functions.
There's even a neat solution method for the reduced cubic based on the trig identity 4 cos³ t - 3 cos t - cos 3t = 0. (make the substitution x = m cos t, and rewrite your cubic in this form)
Noesis
Apr30-06, 04:17 PM
Slightly off topic, but I never learnt anything about substitutions in algebraic manipulations.
Can anyone direct me to a website, a book, or anything that covers them and their uses?
Sounds like they are powerful. I have only seen them in calculus, your generic old u-substitutions.
Thanks.
Curious3141
Apr30-06, 08:37 PM
I guess it's nitpicky, but I don't think you meant "analytical": you meant solutions in terms of radicals.
Yeah, I meant solution by radicals. I was looking for a "snappy" title, and I've seen "analytical" being used in this exact context before, for e.g. here (http://www.me.gatech.edu/energy/andy_phd/appA.htm).
But if the terminology isn't correct, please feel free to amend the topic title.
There's even a neat solution method for the reduced cubic based on the trig identity 4 cosł t - 3 cos t - cos 3t = 0. (make the substitution x = m cos t, and rewrite your cubic in this form)
Yup, I'm aware of this method, but it's nice to keep the whole thing algebraic. Although when you come to writing out the solutions, it's often easier to work in trig ratios. :smile:
HallsofIvy
May3-06, 07:18 AM
Actually, no, that site does NOT refer to "analytical" solutions, it refers to
"ANAYLYTICAL", whatever that means!
Curious3141
May3-06, 07:24 AM
Actually, no, that site does NOT refer to "analytical" solutions, it refers to
"ANAYLYTICAL", whatever that means!