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### Is There a 'Discriminant' for a Quartic Equation?

```Date: 01/12/2005 at 16:57:52
From: Romain
Subject: Quartic equation

Hi,

My question is about quartic equations.  I know how to find the roots
but I would like to know if it is possible to predict the kinds of
roots only with the coefficients.

For example, in a quadratic equation ax^2 + bx + c = 0, you know if
b^2 - 4ac > 0, there are two real roots.  In a cubic equation ax^3 +
bx^2 + cx + d = 0, you know if Yn^2 > h^2, there is one real root and
two complex ones, etc.

Does something similar exist for the quartic equation?

Thanks, and congratulations on your site.

```

```
Date: 01/12/2005 at 18:00:43
From: Doctor Vogler
Subject: Re: Quartic equation

Hi Romain,

Thanks for writing to Dr. Math.  Yes, it does.  The "types" of roots
of a polynomial has a lot to do with a thing called the "Galois group"
of a polynomial (or of the splitting field of the polynomial).  This
is a concept from abstract algebra.  If you have a basic understanding
of abstract algebra, then I would recommend a book by Dummit and Foote
(called Abstract Algebra) which discusses this very subject in great
detail.

First of all, every polynomial has a discriminant, not just
quadratics.  The discriminant is the first place you look to classify
the types of roots a polynomial has.  The discriminant is defined as
the product of the squares of all of the differences of two roots of
the equation.  (Some definitions also multiply by a power of the leading
coefficient in order to make them polynomials in the coefficients even
when the polynomial isn't monic.  This makes little other difference.)
Of course, if you don't know the roots, that makes it
hard to compute.  But, in fact, there are efficient algorithms to get
a formula, for each degree n, that takes the coefficients of a
degree-n polynomial and gives you the discriminant, and the formula
will always be a rational polynomial in the coefficients of your

http://mathworld.wolfram.com/PolynomialDiscriminant.html

This page also gives the discriminants for degree 2, 3, and 4 polynomials.

Now let's consider.  If there is a double root (a repeated root) of
your polynomial, then the discriminant will be zero.  (And, likewise,
a zero discriminant means that there is a repeated root.)

Now suppose your coefficients are all real numbers, and there happen
to be four real roots (p, q, r, s).  Then the discriminant is

[(p-q)(p-r)(p-s)(q-r)(q-s)(r-s)]^2 > 0.

But what if you have two real roots and two complex roots (p, q, r+si,
r-si)?  Then the discriminant is

(p-q)^2 [(p-r-si)(p-r+si)]^2 [(q-r-si)(q-r+si)]^2 (2si)^2

= (p-q)^2 ((p-r)^2 + s^2)^2 ((q-r)^2 + s^2)^2 * (-4s^2) < 0.

Similarly, if you have four complex roots (p+qi, p-qi, r+si, r-si),
then the discriminant will be positive.

So you may conclude that if the discriminant is negative, then you
have two real roots and two complex roots.

You have to do a little more work to distinguish between four real
roots and four complex roots, but there are various ways of doing that
once you know that the discriminant is positive.  I'd have to work on
coming up with one (probably consulting Dummit and Foote in the
process), so I'll stop here for now and see if this is the kind of
thing you want.

more help, please write back and show me what you have been able to
do, and I will try to offer further suggestions.

- Doctor Vogler, The Math Forum
http://mathforum.org/dr.math/

```

```
Date: 01/12/2005 at 22:12:04
From: Romain
Subject: Quartic equation

Thank you--this is exactly the subject I am interested in.  But what
I am really interested to know is if all the roots are complex or not.
Is there a simple way to determine that?

```

```
Date: 01/13/2005 at 11:12:45
From: Doctor Vogler
Subject: Re: Quartic equation

Hi Romain,

After consulting the text I mentioned, I remembered how resolvents
work.  A resolvent is a helping polynomial.  For example, one helpful
resolvent for the polynomial

(x - p)(x - q)(x - r)(x - s)

is the polynomial

(x - (p + q)(r + s))(x - (p + r)(q + s))(x - (p + s)(q + r)).

It turns out that as long as the new polynomial (just like the
original polynomial) is unchanged by permuting the four original roots
(p, q, r, s), you can solve for the coefficients of the new polynomial
using only the coefficients (not the roots) of the original polynomial.
(For info on how to do this, search for "symmetric polynomials".)

For example, if the original polynomial is irreducible (over the
rational numbers) then just as the discriminant tells you some things
about the roots, how (or whether) the resolvent factors can tell you

So my hope was to come up with a cubic resolvent polynomial which
would have three real roots if the original quartic had four real
roots but which would have only one real root (and two complex) if the
original quartic had four complex roots.  Then you could use your
other formula (for the discriminant of the cubic) to distinguish
between the two cases.

But I was having trouble finding such a resolvent.  Maybe I just
wasn't creative enough, but the ones I came up with generally had
three real roots in both of those two cases (though not always in the
other case that we can already tell apart using the discriminant).

Then I had a different idea.  You see, we're really dealing with real
form the roots take, or their degree, but only whether they are real
or complex.  Well, every real quartic polynomial can factor into two
quadratic factors.  So let's see how it factors.

We want to find t, u, v, and w in terms of a, b, c, and d, so that

(x^2 + tx + u)(x^2 + vx + w) = x^4 + ax^3 + bx^2 + cx + d.

Actually, things work out nicer when we force a = 0 by making the
change-of-variables

y = x + a/4

y^4 + ky^2 + my + n

where

k = b - 3a^2/8
m = c - ab/2 + a^3/8
n = d - ac/4 + ba^2/16 - 3a^4/256.

Now we solve

(y^2 + ty + u)(y^2 + vy + w) = y^4 + ky^2 + my + n

which gives us four equations (since these must be the same polynomial
and not just equal for some y), namely

t + v = 0
u + w + tv = k
tw + uv = m
uw = n.

We can solve these like so:

v = -t
w = k - u - tv = k - u + t^2

Substitute into the third equation and solve for q:

u = (t^3 + kt - m)/(2t).

Then we substitute u and w into the last equation, finally ending up with

t^6 + (2k)t^4 + (k^2 - 4n)t^2 - m^2 = 0.

That means that t^2 is a root of the following resolvent cubic:

z^3 + (2k)z^2 + (k^2 - 4n)z - m^2.

(This is where the change-of-variables paid off.  If we hadn't forced a=0,
then we would have a sixth-degree polynomial with six messy coefficients

So now let's think for a moment about how this applies to my other method.
It would appear that I found precisely the resolvent that I was looking for.
So that makes me wonder what form its roots take, in terms of
the roots of the original quartic.

With a little work, I determined that the creative thought that I
didn't have was to use the three roots

(p - q - r + s)^2
(p - q + r - s)^2
(p + q - r - s)^2

(where p, q, r, and s are the roots of the original quartic) because
those roots (divided by 4) are the roots of the resolvent I ended up
with.  Perhaps the reason I didn't come up with these three roots is
that it is not immediately apparent that permuting p, q, r, and s only
permutes those three expressions.  Perhaps the reason is that I just
wasn't creative enough.

In fact, since each term above is unchanged when you add the same
number to every root, you get the same formula if you let p, q, r,
and s be the roots of

y^4 + ky^2 + my + n.

Now, continuing on, my first (incorrect) inclination was
to say that if we have two pairs of complex roots,
then there is only one way to choose our two quadratics (well, two if
you count switching them), but if we have four real roots, then there
are three ways (or six) to choose them.  (In fact, there is also only
one/two ways if there are two real roots and two complex.)  So that
means we only need to check if the resolvent cubic

z^3 + (2k)z^2 + (k^2 - 4n)z - m^2

has one real root or three, which we can do by checking the
discriminant, as you described.

But it turned out that this cubic has exactly the same discriminant
as the original quartic.  This surprised me, until I realized that in
the case of two pairs of complex roots, mis-matching the pairs results
in a value of t which is pure imaginary, meaning that t^2 is a negative
real root of the cubic polynomial.  (In the two-real/two-complex case,
you'll get one real root and two complex roots of the resolvent cubic.)

But all is not lost!  We just need to be able to tell whether our
resolvent cubic

g(z) = z^3 + (2k)z^2 + (k^2 - 4n)z - m^2

has three positive real roots, or one positive and two negative real roots.
So let's look at our cubic.  Its derivative will be zero when

3z^2 + (4k)z + (k^2 - 4n) = 0.

and we can solve this for z using the quadratic formula.  If there
are no real roots, then that means our cubic has one real root
and no local maximum or minimum (which only happens in the
two-real/two-complex case).  If there are two real roots, then
the smaller one is the z value of the local maximum of the cubic,
and the larger one is the z value of the local minimum of the cubic.
A useful exercise which is well worth doing is to check that the
maximum value g(z) is positive and the minimum value g(z) is
negative exactly when the discriminant of the cubic is positive
(i.e. when the cubic has three real roots).

In our case, the important thing to check is that the z value of
the local maximum (which is between the smaller two roots) is negative.
If the discriminant is positive but the z value of the local
maximum is negative, then you have four complex roots.
If the discriminant is positive but the z value of the local
maximum is positive, then you have four real roots.  If the
discriminant is negative, then you have two real roots and
two complex roots.

You'll find that the z value of the local maximum is

z = -(2/3)k - (1/3)sqrt(k^2 + 12n)

which is negative if and only if

-(2/3)k - (1/3)sqrt(k^2 + 12n) < 0

-(2/3)k < (1/3)sqrt(k^2 + 12n)

-2k < sqrt(k^2 + 12n)

k > 0 or 4k^2 < k^2 + 12n

k > 0 or k^2 < 4n.

I'll leave it up to you to rewrite these variables in terms of
the coefficients of your original polynomial.  That's a
straight-forward but messy task, since you'll end up with
some rather large expressions.  If you got lost in all of
this, or need more help coming up with the formulas, or
and I will try to help.  But before I go, I should like to point
out one other completely different but very nice method
to decide the same thing.

It's called Sturm's Theorem.  See

http://en.wikipedia.org/wiki/Sturm's_theorem

for a very simple algorithm to count the number of distinct real roots

f0(x) = f(x)

and its derivative

f1(x) = f'(x).

Then you do polynomial division to get the next polynomial

f0(x)/f1(x) = quotient(x) + remainder(x)/f1(x)

or

f0(x) = quotient(x)*f1(x) + remainder(x)

where the remainder polynomial has to have smaller degree than the
divisor f1(x) (otherwise you chose the quotient incorrectly).  You
call the remainder -f2(x), so that

f0(x) = (polynomial)*f1(x) - f2(x)

and then you repeat

f0(x) = (polynomial)*f1(x) - f2(x)
f1(x) = (polynomial)*f2(x) - f3(x)
f2(x) = (polynomial)*f3(x) - f4(x)
f3(x) = (polynomial)*f4(x) - f5(x)

and so on, until you get a constant.  Then you can stop.  Now to find
out the number of distinct real roots between a and b, you evaluate
all of the polynomials f0, f1, f2, etc. at a and count the number of
times the sign changes.  Then you do the same at b and subtract the
second number from the first.  The result is the number of real roots
between a and b.  To get the total number of real roots, evaluate at
-infinity and +infinity.  Evaluating at infinity means to take the
signs of the leading coefficients of the polynomials.  Evaluating at
-infinity means to take the signs of the leading coefficients of the
polynomials times (-1)^degree.

For example, the polynomial x^6 + 2x^4 + 2x^3 + 2x^2 + 4 yields the
sequence

f0 = x^6 + 2x^4 + 2x^3 + 2x^2 + 4
f1 = 6x^5 + 8x^3 + 6x^2 + 4x
f2 = (-2/3)x^4 - x^3 - (4/3)x^2 - 4
f3 = (-19/2)x^3 - 24x^2 + 32x - 54
f4 = (1916/361)x^2 - (2200/361)x + 2848/361
f5 = (-1385518/229441)x + 484462/229441
f6 = -8517767684/1329404521

Evaluating at -infinity, the signs go

f0 = +
f1 = -
f2 = -
f3 = +
f4 = +
f5 = +
f6 = -

and we see three sign changes.  Evaluating at +infinity, the signs go

f0 = +
f1 = +
f2 = -
f3 = -
f4 = +
f5 = -
f6 = -

and again we see three sign changes.  Subtracting, 3 - 3 = 0, and so
there are no real roots.

- Doctor Vogler, The Math Forum
http://mathforum.org/dr.math/
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Associated Topics:
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