Formula for Finding RootsDate: Fri, 20 Jan 1995 16:48:45 -0600 From: Betty Thompson Subject: formulas for roots I need to know the formula for finding the roots of any quadratic, cubic or quartic polynomial. Also who is credited with the discovery of the last two formulas. Betty Thompson thompson_be@mailserver.capnet.state.tx.us Date: 31 Jan 1995 01:48:35 GMT From: Dr. Ethan Subject: Re: formulas for roots Here are a few answers; but for the quartic, I think I'm just going to send you to the library (I expect that you have a card). This isn't just to give you researching practice, it is also because the method for solving the quartic is extremely long and involved and to type it in would be a nightmare. For quadratic, you can use the much loved quadratic formula. Given ax^2 + bx +c = 0 the roots are x = (-b +- (b^2 - 4ac)^.5)/2a For cubics, things get a little nastier. Here is an explanation given by one of my buddies, Ken, Dr. Stud: "Okay, here's how you do it. Let's say you have the equation ax^3 + bx^2 + cx + d = 0. The first thing you do is to get rid of the a out in front by dividing the whole equation by it. Then we get something in the form of x^3 + ex^2 + fx + g = 0. The next thing we do is to get rid of the x^2 term by replacing x with (x - e/3). That will give us something of the form of x^3 + px + q = 0. This is much easier to solve, although it's still hard. Now introduce two new variables, t and u, defined by u - t = q and tu = (p/3)^3. Then x = CubeRoot{t} - CubeRoot{u} will be a solution of x^3 + px + q = 0. Verify this result now, and make sure you see why it works. To find the other two solutions (if there are any) we could divide x^3 + px + q by its known factor (x - CubeRoot{t} + CubeRoot{u}), getting a quadratic equation that we could solve by the quadratic formula. So that's the basic idea behind the cubic. If you wanted to find the actual expression for t and u in terms of p and q, you could solve those two equations defining p and q (substitution would probably be easiest). Then you could obtain a real formula for x in terms of p and q." As you can see the complexity has greatly increased. To try to go backwards and come up with a closed form in the original a, b, c, d would be a real pain. Well the quartics are even worse. But you should be able to find them in most modern algebra texts.(And if you happen to find one that doesn't give them, then they certainly will cite a place where they can be found.) Hope this helps, Ethan Doctor On Call |
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