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Roots of Cubic Equations


Date: 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
    
Associated Topics:
College Modern Algebra

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