Solving Quartic Equations - Up to 4 Solutions

[Mentallic]

• 
  A linear equation is quite simple to solve, up to 1 solution can be obtained.

$$ax+b=0$$

$$x=\frac{-b}{a}$$

• A quadratic equation can have up to 2 solutions, by using $$\pm$$ from taking the root of the square.

$$ax^{2}+bx+c=0$$

$$x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$$

• A cubic can have 3 solutions.

Now my question is, since I have seen a formula to solve cubics (even though attempting to use it can be an eerie thought) how is it possible to obtain 3 solutions from that 1 formula?

• A quartic can have 4 solutions by using 2 $$\pm$$ in the formula

 

[dodo]

Nice question, I learned something while figuring out an (incomplete) answer. :) Which is:

Because there are cubic roots in the formula. From the Wikipedia page for "Cubic root",

"All real numbers have exactly one real cube root and a pair of complex conjugate roots."

Now, there are two cubic roots in the cubic formula, but they are of the form cubicroot(a+b) and cubicroot(a-b). I'd presume/imagine/I-don't-know that the six combinations reduce to three given the symmetry of the two arguments.

 

[Mentallic]

I have yet to study complex numbers, so the conjugates of those aren't really in my reach of knowledge.

However, from what you've said, I think it may be that the roots are expressed in the part of the formula:

$$x=\sqrt[3]{m+\sqrt{n}}+\sqrt[3]{m-\sqrt{n}}$$ where m & n are complex (and I don't necessarily mean imaginary)

Is it possible the value of n in both square roots determine the last 2 roots of the cubic?
Say, if:
n<0 (no real roots for that pair)
n=0 (2 equal roots)
n>0 (2 different roots)

If my statement was true, then I'm still having a problem of understanding how the 3rd (always real) root is derived from the http://www.math.vanderbilt.edu/~schectex/courses/cubic/"

 

[robert Ihnot]

One root would have to be real because 3 is an odd number and can be positive or negative, and the leading term will dominate the equation. (This, of course, means that the coefficients of the equation are all real, a normal assumption.) With imaginary coefficients it could be different, such as (x-i)^3.

The fact is there is the "Irreducible Cubic," where all roots are real, but they appear as imaginary. You use DeMovier’s formula to take care of that. (Cardan, himself, noticed the case of X^3=15X+4, where 4 is a solution.)

On the other hand, if two complex roots, then they are always conjugates; so that in multiplying them, the imaginary part goes out. (X-(a+bi))(X-(a-ib))=X^2-2aX+a^2+b^2.

The formula you give, $$x=\sqrt[3]{m+\sqrt{n}}+\sqrt[3]{m-\sqrt{n}}$$ is only good for one root, unless we remember the other roots involve the use of the cubic root of 1. Which is a root of the same equations. Because the cubic root involves three possible solutions. (But, there are actually more solutions to the above equation, but more than three are reduntant.)

If you have not studied complex numbers, then this would be confusing. One author says this about the above formula: "I do not recommend that you memorize these formulas. Aside from the fact that it's too complicated, there are other reasons why we don't teach this formula to calculus students. One reason is that we're trying to avoid teaching them about complex numbers. Complex numbers (i.e., treating points on the plane as numbers) are a more advanced topic, best left for a more advanced course. But then the only numbers we're allowed to use in calculus are real numbers (i.e., the points on the line)." http://www.math.vanderbilt.edu/~schectex/courses/cubic/

 

[Mentallic]

I have been exposed to imaginary numbers somewhat, due to studying parabolas and the like.
The author also adds:

... That problem has real coefficients, and it has three real roots for its answers. ... Ultimately, the square roots of negative numbers would cancel out later in the computation, but that computation can't be understood by a calculus student without additional discussion of complex numbers.

I can understand this, but would like to know - if possible - where exactly these complex conjugates are located in the formula; also, how these 2 roots are added as a solution to the last real root.

 

[Integral]

Once you have a single real root you can factor your cubic leaving a quadratic. Now apply the quadratic formula, this will give you 2 more roots, either 2 real roots or a complex conjugate pair.

 

[Mentallic]

Yes it is quite obvious through factorization, but often the first root isn't given and can't be determined graphically since it is irrational. Somehow the 3 roots are obtained from that formula and I can't seem to come to terms with how it is done so (and where it happens specifically).

 

[robert Ihnot]

In the reduced equation X^3+pX+q = 0. $$\sqrt{n}=\sqrt{(p/3)^3+(q/2)^2}$$ So that if the terms under the square root is negative then n is a complex number. And you have to find a way to extract that from the cubic roots. An example is Cardan's own example: X^3-15X+-4 = 0, yielding n=-121.
Of course, as pointed out by Integral and later by quadraphonics, it is simpler to just factor out X-4.

 

[quadraphonics]

Yes it is quite obvious through factorization, but often the first root isn't given and can't be determined graphically since it is irrational. Somehow the 3 roots are obtained from that formula and I can't seem to come to terms with how it is done so (and where it happens specifically).

It's obtained by first located a real root, and then factorizing and applying the quadratic formula. The real root is obtained by "depressing" the cubic:

http://www.sosmath.com/algebra/factor/fac11/fac11.html

 

[Mentallic]

It's obtained by first located a real root, and then factorizing and applying the quadratic formula. The real root is obtained by "depressing" the cubic:

http://www.sosmath.com/algebra/factor/fac11/fac11.html

Ahh this makes perfect sense now

So when the cubic formula is used, it gives the real root z. From there simply factorize the equation to obtain the quadratic and use the quadratic formula to obtain the last 2 conjugate pairs.

Can it be determined graphically or through some simple algebra which the real root of the cubic is going to be (the root which using the cubic formula will lead one to)?

 

[robert Ihnot]

Mentallic: So when the cubic formula is used, it gives the real root z. From there simply factorize the equation to obtain the quadratic and use the quadratic formula to obtain the last 2 conjugate pairs.

Sometimes there is three real roots.

Can it be determined graphically or through some simple algebra which the real root of the cubic is going to be (the root which using the cubic formula will lead one to)?

You can always draw the graph, especially if only one root. In Cardan's own example, where we have three real roots: 4, -2+sqrt3, -2-sqrt3. If we work out the details not using the cube root of 1, we finally arrive at $$Y=\sqrt{5}*2*cos(26.57)$$ which works out, should be exact with enough details, to 4.

 

[Mentallic]

Sometimes there is three real roots.

Yes then the conjugates (the quadratic factor) will include 2 more reals to the cubic.
The cubic always has 1 real and the other 2 roots are determined by the nature of the quadratic.

I really hope I'm on the right track here.

 

[Ben Niehoff]

To understand how to treat the two cube roots, you need to understand how they were derived in the first place. It turns out that they are reciprocals of each other (up to the cube root of a rational factor). In other words, they are conjugates, in the following sense:

$$\frac{1}{\sqrt[3]{R + \sqrt{S}}} = \frac{1}{\sqrt[3]{R + \sqrt{S}}} \frac{\sqrt[3]{R - \sqrt{S}}}{\sqrt[3]{R - \sqrt{S}}} = \frac{\sqrt[3]{R - \sqrt{S}}}{\sqrt[3]{R^2 - S}}$$

Notice how multiplication by the conjugate eliminates the nested square root.

Then the full cubic formula can be written

$$x = P + Q + Q^*$$

where * is algebraic conjugation as shown above, and

$$Q = \sqrt[3]{R + \sqrt{S}}$$

To find the other two cube roots, we have to multiply by any of the three complex cube roots of 1:

$$\omega \in \left \{ 1, \; -\frac12 + i \frac{\sqrt3}2, \; -\frac12 - i \frac{\sqrt3}2 \right \}$$

which gives

$$x = P + \omega \sqrt[3]{R + \sqrt{S}} + \left(\omega \sqrt[3]{R + \sqrt{S}} \right)^*$$

and finally

$$x = P + \omega \sqrt[3]{R + \sqrt{S}} + \omega^* \sqrt[3]{R - \sqrt{S}}$$

where now * is complex conjugation, in order to give the same conjugation relationship written at the beginning of this post. This equation specifies exactly how to combine the three possible cube roots to give exactly three values for x, rather than six. Also, one can see that

1. If S is negative, then the second radical is always the complex conjugate of the first, and so there are three distinct real roots,

2. If S is positive, then there will be one real root (for $\omega = 1$) and two complex conjugate roots,

3. If S is zero, then two roots will be the same (and all real), and

4. If R is zero and S is zero, then all three roots will be the same.