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### Do Irrational Roots Always Occur in Conjugate Pairs?

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Date: 03/13/2002 at 08:51:29
From: Libby Lerner
Subject: Do irrational roots always occur in conjugate pairs?

A colleague and I have been puzzled by the equation x^3+4x^2+3x+1 = 0.
Upon graphing, there appear to be two imaginary solutions and one real
solution. According to the rational root theorem, if the root is
rational, then it should be either 1 or -1. It isn't. If the root is
irrational, then its conjugate should also be a root. It isn't, since
there is only one real root. What are we missing?

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Date: 03/13/2002 at 11:17:20
From: Doctor Paul
Subject: Re: Do irrational roots always occur in conjugate pairs?

The irrational conjugate roots theorem says:

Let p(x) be any polynomial with rational coefficients. If
a + b*sqrt(c) is a root of p(x), where sqrt(c) is irrational and
a and b are rational, then another root is a - b*sqrt(c).

The problem is that you have no guarantee that the root of a generic
cubic can be written in the form x = a + b*sqrt(c).

It is in fact the case that whenever a cubic has only one real
irrational root, this irrational root will never be expressable in the
form x = a + b*sqrt(c), where a and b are rational and sqrt(c) is
irrational. If this were not the case, then the irrational conjugate
roots theorem would tell us that x = a - b*sqrt(c) would be another
irrational root, and we would have problems (we are assuming there is
only one real root) unless we could conlude that a + b*sqrt(c) =
a - b*sqrt(c) (so that we in fact had only one real root). But if
a + b*sqrt(c) = a - b*sqrt(c), then 2*b*sqrt(c) = 0, which forces
b = 0 (since sqrt(c) is irrational, c cannot be zero). But b = 0 means
that the root is just x = a, where a is rational. And this cannot be,
since we assumed that the root was irrational.

Thus it must be the case that whenever a cubic has only one real
irrational root, this irrational root will never be expressable in the
form x = a + b*sqrt(c) where a and b are rational and sqrt(c) is
irrational.

Notice that if the cubic has one real rational root, say x = p/q then
choose a = p/q, b = 0, and c = any nonsquare to satisfy the irrational
conjugate roots theorem.  In this case, a + b*sqrt(c) = a =
a - b*sqrt(c), and things work out nicely.

Let's look more specifically at what's going on with

p(x) = x^3+4x^2+3x+1

We need to solve p(x) = 0, and I'll use Cardano's formula, as
described in our FAQ:

Cubic and Quartic Equations
http://mathforum.com/dr.math/faq/faq.cubic.equations.html

Make the substitution x = (y - 4/3) to obtain:

y^3- 7/3*y + 47/27  (verify this)

so if we have

y^3 + p*y + q = 0

then it must be the case that

p = -7/3 and q = 47/27

Now I deviate from the FAQ a bit (I like my method better), but the
ideas are similar.

Let D = -4*p^3 - 27*q^2 (this is the discriminant of the cubic)

Then D = -4 * (-7/3)^3 - 27 * (47/27)^2 = -31

Then we let

[ -27 * q      3 * sqrt(-3 * D) ]
A = cuberoot |---------  + ----------------- |
[    2               2          ]

and

[ -27 * q      3 * sqrt(-3 * D) ]
B = cuberoot |---------  - ----------------- |
[    2               2          ]

So in this case,

A = [-47/2 + 3/2*sqrt(93)]^(1/3)

and

B = [-47/2 - 3/2*sqrt(93)]^(1/3)

Then

L = (A+B)/3

M = (t^2*A + t*B)/3

and

N = (t*A + t^2*B)/3

are the three roots, where t is the first primitive third root of
unity:

t = -1/2 + i*sqrt(3)/2

Thus we have:

1            [ -47     3*sqrt(93) ]
L =    - * cuberoot |-----  + ---------- | +
3            [  2          2      ]

1            [ -47     3*sqrt(93) ]
- * cuberoot |-----  - ---------- |
3            [   2         2      ]

It is a simple exercise to verify that y^3- 7/3*y + 47/27 has only one
real root. Inasmuch as L ~= -1.814565703 is real, we have found the
only real root of y^3- 7/3*y + 47/27 = 0. (I am not computing M and N
because they're messy and they're not really needed to talk about what
we're discussing)

Thus L' = L - (4/3) ~= -3.14789903633 (remember we made the
substitution x = y - 4/3) will satisfy the original cubic p(x).

Thus we have:

1            [ -47     3*sqrt(93) ]
L' =   - * cuberoot |-----  + ---------- | +
3            [  2          2      ]

1            [ -47     3*sqrt(93) ]   4
- * cuberoot |-----  - ---------- | - -
3            [   2         2      ]   3

with p(L') = 0.

As determined earlier, L' will not be expressible in the form
a + b*sqrt(c), and so the irrational conjugate roots theorem does not
apply.

I suppose there is a more direct way to prove that L' will not be
expressible in the form a + b*sqrt(c) without relying on the
irrational conjugate roots theorem, but it's not immediately clear to
me how this would be done.

Here's another example I think you might like. Consider the
polynomial:

p(x) = x^3 + x^2 - 2

Plainly, x = 1 is a root and dividing p(x) by (x-1) allows us to use
the quadratic formula to find out that the other two roots are
x = -1 +- i

But using Cardano's formula as described above gives us the following
real root:

1/3 * [ cuberoot(26 + 15*sqrt(3)) + cuberoot(26 - 15*sqrt(3)) - 1]

Unless we want to start being contradictory, this real root had better
be equal to one. And that is in fact the case, but seeing this
requires us to be a bit sneaky. Notice that:

cuberoot(26 + 15*sqrt(3)) = 2 + sqrt(3)

and that

cuberoot(26 - 15*sqrt(3)) = 2 - sqrt(3)

thus

cuberoot(26 + 15*sqrt(3)) + cuberoot(26 - 15*sqrt(3)) =

2 + sqrt(3) + 2 - sqrt(3) = 4

and the root is 1/3 * (4 - 1) = 1.

How do we see that cuberoot(26 + 15*sqrt(3)) = 2 + sqrt(3)?

26 + 15*sqrt(3) = 26 + 15*sqrt(3)

= 8 + 12*sqrt(3) + 18 + 3*sqrt(3)

=   2^3
+ 3 * 2^2 * sqrt(3)^1
+ 3 * 2^1 * sqrt(3)^2
+ sqrt(3)^3

= (2 + sqrt(3))^3

taking cuberoots on both sides yields the desired result:

cuberoot(26 + 15*sqrt(3)) = 2 + sqrt(3)

So in this case the real root of p(x) was rational and we were able to
express our cuberoot terms in the proper form.  I think that if you
investigate this further, you'll find that this is not a coincidence.

some more or if anything is unclear. I've left out some steps to save
time. If you have trouble reconstructing my work, just let me know and
I'll go back and fill in the details.

- Doctor Paul, The Math Forum
http://mathforum.org/dr.math/
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Associated Topics:
High School Exponents
High School Polynomials

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