Roots of Cubic EquationsDate: 9 Jul 1995 00:32:00 -0400 From: Luke Moore Subject: Third degree equations Hello Dr. Math! I'm looking for equations which will tell me how many real roots a third degree equation in the form ax^3 + bx^2 + cx + d = 0 has, and what they are. (I've finished the three OAC maths offered at the high school level, so you have some idea of where I'm at.) Luke Date: 9 Jul 1995 15:56:35 -0400 From: Dr. Ken Subject: Re: Third degree equations Hello there! 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." To try to go backwards and come up with a closed form in the original a, b, c, d would be a real pain, but it could be done if you want a big yucky formula. -K Date: 1/29/96 at 22:28:13 From: John Vogler Subject: Math Stuff Dear Dr. Math, One of the first problems that I had when I heard of Dr. Math was the cubic formula (the other was the Proof of Euler's Theorem). I was thrilled to find it already answered. I started playing around with it, and it always worked perfectly algebraically, but I could never get the right answer out, even when I knew what it was already. The problem was mainly that I was trying to evaluate the real sum of the cubic roots of two complex numbers. That is, (a+bi)^(1/3) + (a-bi)^(1/3). It worked algebraically, but it wasn't very nice for my calculator (which could only do the four functions on complex numbers). Then I figured it out. Here's what you said: Start with ax^3 + bx^2 + cx + d = 0 Divide by a, get x^3 + ex^2 + fx + g = 0 Substitute u = x + e/3, get u^3 + pu + q = 0 (where p = f - (e^2)/3 and q = 2(e^3)/27 - ef/3 + g) Though I have found that it is easier to work with u^3 - 3pu - 2q = 0 (where p = (e^2)/9 - f/3 and q = ef/6 - (e^3)/27 - g/2) Find (r, s) such that r + s = 2q and rs = p^3 (r = q + (q^2 - p^3)^(1/2), and s = q - (q^2 - p^3)^(1/2)) Then, u = r^(1/3) + s^(1/3) and x = u - e/3 The problem is that always when there are three roots, and sometimes when only one, r and s are complex conjugates. Their cube roots are as well, so the sum is real, but that makes it no easier on a simple calculator. So, after a bit of solving, I ended up with this formula: (Use when p < 0.) u = 2 p^(1/2) cos ((1/3) arccos (q)p^(-3/2)) I can't simplify the cos ((1/3) arccos x), but that's a trisected angle, so I rather expect that it can't be simplified except to give the original third degree equation. Meanwhile, this one is nice to your calculator. (The proof deals primarily with the fact that cos 3x = 4(cos x)^3 - 3(cos x).) Hope it helps, Johnny Vogler |
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