The cubic formula

My son and I were trying to use the cubic formula the other day to find the roots of a cubic equation. It’s not as simple as the quadradic formula:

    \[ \begin{array} {*{20}c} {x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} & {{\rm{when}}} & {ax^2 + bx + c = 0} \\ \end{array}\]

We used the method where you first convert the cubic into a depressed cubic. A depressed cubic is missing the x^2 term. It looks like this:

    \[ \begin{array} {*{20}c} {ax^3 + cx + d = 0} \\ \end{array}\]

You convert a normal cubic into a depressed cubic with a substitution. Starting with a general cubic polynomial:

    \[ \begin{array} {*{20}c} {ax^3 + bx^2 + cx + d = 0} \\ \end{array}\]

Make the substitution:

    \[ \begin{array} {*{20}c} {x = y - \frac{b}{3a}} \\ \end{array}\]

And you end up with the depressed cubic:

    \[ \begin{array} {*{20}c} {ay^3 + (c - \frac{b^2}{3a})y + (d + \frac{2b^3}{27a^2} - \frac{bc}{3a}) = 0} \\ \end{array}\]

Finally you find the roots for the depressed cubic to solve for y and substitute back to find the roots for the original equation. Solving for the depressed cubic is a multi-step operation, but it’s not too difficult. The above link explains it nicely.

To make it easier, we decided to try to solve a cubic that was already depressed. We picked the following because one of the three roots (x = 1) is obvious.

    \[ \begin{array} {*{20}c} {x^3 + x - 2 = 0} \\ \end{array}\]

The other two roots are easy to discover. Since (x = 1) is a root, we know {x^3 + x - 2 = 0} can be expressed {(x - 1)P = 0} where P is a quadratic expression. Long division yields {(x^3 + x - 2) = (x - 1)(x^2 + x + 2)} which tells us the other two roots are {x = \frac{-1 \pm \sqrt{-7}}{2}}.

So now we know what to expect.

When you solve this depressed quadratic you get:

    \[ \begin{array} {*{20}c} {x_1 = \sqrt[3]{\sqrt{\frac{28}{27}} + 1} - \sqrt[3]{\sqrt{\frac{28}{27}} - 1}}\\ \end{array}\]

If you put it thru a calculator, you’ll find the above expression indeed equals 1. I was a little surprised to find two irrationals separated by an integer in a non-obvious way. I also found it unsatisfying, since I could not simplify this expression all the way to 1 using algebra.

Let’s look at another cubic solution. Wikipedia tells us the following formula will give us the real root:

    \[ \begin{array} {rl} {x_1 = } & {- \frac{b}{3a}} \\ {} & {- {\frac{1}{3a}}\sqrt[3]{\frac{1}{2}\left(Q + \sqrt{Q^2 - 4(b^2 - 3ac)^3}\right)}} \\ {} & {- {\frac{1}{3a}}\sqrt[3]{\frac{1}{2}\left(Q - \sqrt{Q^2 - 4(b^2 - 3ac)^3}\right)}} \\ \end{array}\]

where

    \[ \begin{array} {rl} {Q =} & {2b^3 - 9abc + 27a^2d} \\ \end{array}\]

To solve {x^3 + x - 2} = 0 we plug (a = 1), (b = 0), (c = 1), and (d = -2) into the above to get:

    \[ \begin{array} {rl} {x_1 =} & {- \frac{1}{3}\sqrt[3]{\frac{1}{2}\left[-54 + \sqrt{(54^2 + 4*27)}\right]}} \\ {} & {- \frac{1}{3}\sqrt[3]{\frac{1}{2}\left[-54 - \sqrt{(54^2 + 4*27)}\right]}} \\ \end{array}\]

which simplifies to

    \[ \begin{array} {*{20}c} {x_1 = - \frac{1}{3}\left[\sqrt[3]{-27 + 6\sqrt{21}} + \sqrt[3]{-27 - 6\sqrt{21}}\right]} \\ \end{array}\]

and finally evaluates (with a calculator) to:

    \[ \begin{array} {*{20}c} {x_1 = 1}\\ \end{array}\]

Although that’s interesting, it’s no more satisfying than the first solution. I feel I’m missing something since I can’t take it all the way to (x=1) algebraically. It was still a fun exercise though. I’ve occasionally wondered about the methods for solving cubic equations, and now I’ve finally seen it done.

I can see why they don’t teach this in high school. It’s too complicated to ask people to memorize, so they wouldn’t want it on a test. And it gives funny looking results, full of square- and cube-roots, that are difficult to manipulate.

Hmmm, now I’m wondering about a general method for finding the roots of a 4th-order equation. Or even a for a 5th-order equation.


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Comments

7 Responses to “The cubic formula”

  1. Z1 guy on October 25th, 2010 6:43 pm

    Great detailed info, I just saved you on my google reader.

    Sent from my iPad 4G

  2. Peter Luthy on November 3rd, 2010 12:01 am

    Hi Neal,

    I agree with you that this formula is too complicated to expect anyone to remember. Even the method for solving it is already somewhat complicated whereas the idea behind the quadratic equation (completing the square) is simple and easy to understand.

    As far as your question about the quartic and quintic equations is concerned, the “quartic equation” exists but is terrifically, horribly complicated. As I recall it takes several pages to write down. So you might expect that the situation with the quintic is mind-bogglingly complicated. But in fact it is even more interesting than that: it is impossible! More precisely, one cannot find the roots of a quintic polynomials in any analogous way, i.e. using normal algebra and square roots, cube roots, fourth roots, etc. This is typically proved in an undergraduate-level course as an application of something called Galois theory which is quite general and abstract, but is much less-complicated sounding when one works with more concrete things like polynomials. Evariste Galois is an interesting historical character in his own right, being a young genius mathematician who was killed in a pistol duel and hastily wrote his ideas down in the days preceding his death: http://en.wikipedia.org/wiki/Évariste_Galois

  3. Neal on November 3rd, 2010 3:14 pm

    Peter, thanks for pointing me at Galois. What a genius, what a sad story. I don’t know much about group theory, but now I’m determined to learn a little more.

    It’s nice being an amateur — you can enjoy something without worrying about being good at it.

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