x^3 + a x^2 + b x + c = (x-a)(x-b)(x-c)
Harry J. Smith
that have its roots equal to its coefficients.
x^3 + a x^2 + b x + c = (x-a)(x-b)(x-c)
Can you list some or all of the solutions?
How many solutions are there?
-Harry
Richard Thomas
>
> Can you list some or all of the solutions?
> How many solutions are there?
Equate co-efficients:
x^3 + a*x^2 + b*x + c = x^3 - (a+b+c)*x^2 + (ab+ac+bc)*x - abc
(1) 2a + b + c = 0
(2) b = ab + ac + bc
(3) c + abc = 0
(3) Gives, c = 0 or
ab = -1
Solving for c = 0 we get pairs of (a,b) = (1,-2) or (0,0) two solutions of
the equation.
Solving for ab = -1 gives us the following system of equations:
b^4 - b^3 - 2b - 2 = 0
c = (2/b) - b
a = -(1/b)
The quartic has only two solutions, b = -0.70348 (5 s.f.)
b = 1.8737 (5 s.f.)
Thus the following sets (a,b,c) are solutions:
(1,-2,0)
(0,0,0)
(1.42,-0.70,-2.14)
(0.53,1.87,-0.81)
Nat Silver
> x^3 + a x^2 + b x + c = (x-a)(x-b)(x-c)
> Can you list some or all of the solutions?
> How many solutions are there?
We know polynomials factor uniquely.
We know its coefficents are symmetric
functions of the roots. That is,
1) a = -(a+b+c)
2) b = ab+ac+bc
3) c = -abc
From 3, we get c(1+ab) = 0.
So c = 0 or ab = -1.
If c = 0, then from 2, we have
b(1-a) = 0.
If b = 0, from 1, we get 2a = 0.
So, a = b = c = 0.
If c is not zero, then ab = -1.
And neither a nor b is zero.
From 2, we have: b = -1 + ac+bc.
From 1, we get: b+c = -2a.
Solving for the coefficients in terms
of a gives:
a = a
b = -1/a
c = (-2a^2 + 1)/a
So, a is not equal to +1/sqrt(2) or -1/sqrt(2).
Solving for the coefficients in terms
of b gives:
a = -1/b
b = b
c = (2 - b^2)/b
So, b is not equal to +sqrt(2) or -sqrt(2).
Solving for the coefficients in terms
of c gives:
If c is not zero, then ab = -1.
So, neither a nor b is zero.
From 2, we have: b = -1 + ac+bc.
-1/a = 1 + ac - c/a
ca^2 + a + (1-c) = 0, which gives
a = (-1+-abs(2c-1))/2c
b = 2c/(1-+abs(2c-1))
c = c.
If c = 1, then a = -1, b = 1.
In summary, all coefficients are zero, or
If c = 1, then a = -1, b = 1.
Nat Silver
> Harry J. Smith wrote:
> > x^3 + a x^2 + b x + c = (x-a)(x-b)(x-c)
> > Can you list some or all of the solutions?
> > How many solutions are there?
>
> We know polynomials factor uniquely.
> We know its coefficents are symmetric
> functions of the roots. That is,
> 1) a = -(a+b+c)
> 2) b = ab+ac+bc
> 3) c = -abc
>
> From 3, we get c(1+ab) = 0.
> So c = 0 or ab = -1.
>
> If c = 0, then from 2, we have
> b(1-a) = 0.
> If b = 0, from 1, we get 2a = 0.
> So, a = b = c = 0.
so we get a = 1, b = -2, c = 0.
In summary, all coefficients may be zero.
We also have a = 1, b = -2, c = 0.
If c = 1, then a = -1, b = 1.
a is not equal to +1/sqrt(2) or -1/sqrt(2).
b is not equal to +sqrt(2) or -sqrt(2).
Otherwise the parameters, give the solutions.
Virgil
If this equation is to be true for infinitely many real x, then
a,b, and c must satisfy the equations:
a + b + c = -a,
a b + b c + c a = b, and
a b c = -c.
Solving this system of equations appears to give 4 real solutions, 3
rational solutions and one irrational solution.
The rational solutions are
(a,b,c) = (0,0,0)
(a,b,c) = (1,-2,0) and
(a,b,c) = (1,-1,-1)
For the irrational solution, (a,b,c), b is the real zero of the
cubic polynomial x^3 - 2 x + 2, i.e.,
b = curt(-1 + sqrt(19/27)) + curt(-1 - sqrt(19/27)),
where sqrt indicates the positive square root function and curt
indicates the real cube root function, and then a = -1/b and
c = 2 a - b.
This last solution is approximately given by
(a,b,c) ~ (.565197717384, -1.76929235424, .63889619471)
Zdislav V. Kovarik
After seeing three answers, which partially checked
and partially didn't, it became clear that this is
a good exercise on patience and consistency. After a
first, careless and failed, attempt, I concentrated
better and found:
three integer-valued answers,
one more real answer with irrational entries,
two more complex answers (complex conjugate),
no more answers over the complex scalars,
more possible answers over fields of characteristic 2
(namely (a, b, c)= (a, 0, 0), a arbitrary).
They all checked.
(And it is a convention to say that a cubic
polynomial has degree 3, rather than order 3.)
Cheers, ZVK(Slavek).
Harry J. Smith
"Zdislav V. Kovarik" <kov...@mcmail.cis.mcmaster.ca> wrote
in message
news:Pine.SOL.4.33.02072...@mcmail.cis.mcma
ster.ca...
> (Did I see it in a book of competition problems?)
>
> polynomial has degree 3, rather than order 3.)
I agree, but I was using the termonology I got from
mathworld at:
http://mathworld.wolfram.com/Polynomial.html
They use both order and degree at:
http://mathworld.wolfram.com/OrderPolynomial.html
and
http://mathworld.wolfram.com/DegreePolynomial.html
degree is preferred.
> Cheers, ZVK(Slavek).
-Harry
Harry J. Smith
"Virgil" <vmh...@attbi.com> wrote in message
news:vmhjr2-8A59AE....@netnews.attbi.com...
Typo: c = -2 a - b.
> This last solution is approximately given by
> (a,b,c) ~ (.565197717384, -1.76929235424, .63889619471)
FWIW, I agree that these four are the only real solutions.
-Harry
Harry J. Smith
"Richard Thomas" <chardt...@icqmail.com> wrote in message
news:ai1nco$s54$1...@newsreaderg1.core.theplanet.net...
The last two solutions are incorrect because the quartic
should be:
b^4 + b^3 - 2b^2 + 2 = 0.
By inspection we can see that b = -1 is a root, so (a,b,c) =
(1, -1, -1) is a solution.
Divide the quartic by b+1 and we get b^3 - 2 b + 2 = 0. See
Virgil's post for the real root of this, and the final real
solution.
-Harry
Harry J. Smith
"Nat Silver" <mat...@worldnet.att.net> wrote in message
news:OtZ09.11489$Kl6.6...@bgtnsc04-news.ops.worldnet.att.n
et...
No, (a,b,c) = (-1, 1, 1) is not a solution.
(a,b,c) = (1, -1, -1) is a solution.
> a is not equal to +1/sqrt(2) or -1/sqrt(2).
> b is not equal to +sqrt(2) or -sqrt(2).
> Otherwise the parameters, give the solutions.
See Virgil's post for the final real solution.
-Harry
Nat Silver
> Zdislav V. Kovarik wrote:
> > (And it is a convention to say that a cubic
> > polynomial has degree 3, rather than order 3.)
> I agree, but I was using the termonology I got from
> mathworld at:
1. "Mathworld" does not set the standard for math terminology.
Although a convenient and excellent source, Eric inevitably has
made many errors, some of which have been noted by sci.math.
>degree is preferred.
yes.
2. Would you please give us the source of your problem?
> > (Did I see it in a book of competition problems?)
3. Note: Virgil used a computer algebra system.
Virgil wrote:
The rational solutions are
(a,b,c) = (0,0,0)
(a,b,c) = (1,-2,0) and
(a,b,c) = (1,-1,-1)
For the irrational solution, (a,b,c), b is the real zero of the
cubic polynomial x^3 - 2 x + 2, i.e.,
b = curt(-1 + sqrt(19/27)) + curt(-1 - sqrt(19/27)),
where sqrt indicates the positive square root function and curt
indicates the real cube root function, and then a = -1/b and
c = 2 a - b.
This last solution is approximately given by
Virgil
<IPh19.9584$pg2.7...@bgtnsc05-news.ops.worldnet.att.net>,
"Nat Silver" <mat...@worldnet.att.net> wrote:
> Harry J. Smith wrote:
> > Zdislav V. Kovarik wrote:
> > > (And it is a convention to say that a cubic
> > > polynomial has degree 3, rather than order 3.)
> > I agree, but I was using the termonology I got from
> > mathworld at:
>
> 1. "Mathworld" does not set the standard for math terminology.
> Although a convenient and excellent source, Eric inevitably has
> made many errors, some of which have been noted by sci.math.
>
> >degree is preferred.
>
> yes.
>
> 2. Would you please give us the source of your problem?
>
> > > (Did I see it in a book of competition problems?)
>
> 3. Note: Virgil used a computer algebra system.
In addition to using head, pencil and paper.
Except for evaluating the approximate solution to about 12
significant digits, the problem is not too difficult to do the old
fashioned way. I did it first approximately using the HP49G
calculator, then did it over by hand to verify the result. The only
part not done by hand was the approximation of the irrational
solution. My HP49G will not do the exact zeros of the cubic x^3 - 2
x + 2, so that part was done only by hand.
Harry J. Smith
t...
> Harry J. Smith wrote:
> > Zdislav V. Kovarik wrote:
>
> 2. Would you please give us the source of your problem?
>
> > > (Did I see it in a book of competition problems?)
I do not know the source. I found it in some notes I made in
1954 when I was a student at the University of Dayton in
Ohio.
> 3. Note: Virgil used a computer algebra system.
>
> Virgil wrote:
>
> The rational solutions are
> (a,b,c) = (0,0,0)
> (a,b,c) = (1,-2,0) and
> (a,b,c) = (1,-1,-1)
>
> For the irrational solution, (a,b,c), b is the real zero
of the
> cubic polynomial x^3 - 2 x + 2, i.e.,
> b = curt(-1 + sqrt(19/27)) + curt(-1 - sqrt(19/27)),
> where sqrt indicates the positive square root function and
curt
> indicates the real cube root function, and then a = -1/b
and
> c = 2 a - b.
Typo: Change to c = -2 a - b.
> This last solution is approximately given by
> (a,b,c) ~ (.565197717384, -1.76929235424, .63889619471)
-Harry