Cubic Equation z^3 + Az^2 + Bz + C = 0

[neo_maya]

Guys,

 

How do you solve this one?

 

z^3 + Az^2 + Bz + C = 0 ;

 

So far I have reached this solution ( resources from web of course )

 

w = (-f +- sqrt ( 27f^2 + 4e^3 )/ 27 ) / 2

 

* Substitution :

 

z = y - b/3;

y = x + s/x;

s = -e/3;

w = x^3;

 

 

e = (c - b^2)/3

f = d + (2b^3/27) - (bc)/3;

[matt grime]

google for Cardan, or do i mean cardano?

[psi20]

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

 

Cardano's Ars Magna way back had the solution to this, but it was solved earlier

[Primarygun]

In SOME caseS, if given that one root is an integer, put the prime number of the constant term into the equation ,together with the negative one.

Then you can get a root and by factorizing theorem, factorize it to a qudratic equation.

[uncool]

First thing: get rid of A. divide the equation by A.

Second thing: get rid of B. Do this by a constant shift - y = x - B/3.

Then, let y = z - p/3z where p is the y-coefficient. This will leave you with an equation with z^3, constand, and 1/z^3 terms.

Multiply by z^3.

Let u = z^3.

Solve for u by the quadratic formula. either solution works - they reveal the same final answer.

Solve for z.

Substitute z into the equation for y.

Find x.

Done.

-Uncool-

[Johnny5]

First thing: get rid of A. divide the equation by A.

Second thing: get rid of B. Do this by a constant shift - y = x - B/3.

Then' date=' let y = z - p/3z where p is the y-coefficient. This will leave you with an equation with z^3, constand, and 1/z^3 terms.

Multiply by z^3.

Let u = z^3.

Solve for u by the quadratic formula. either solution works - they reveal the same final answer.

Solve for z.

Substitute z into the equation for y.

Find x.

Done.

-Uncool-[/quote']

 

I needed this just a few days ago, and here it is.

 

Let's see, you want to solve a cubic equation. Here is the general form:

 

[math] a_3 Z^3 + a_2 Z^2+a_1Z+a_0 = 0[/math]

 

Since the equation is necessarily a cubic equation, it is not the case that a_3 is equal to zero, therefore we can divide both sides of the equation by the coefficient of the leading term of the polynomial to obtain the following statement, which has the same roots:

 

[math] z^3 + Az^2 + Bz + C = 0 [/math]

 

Now, you say the first thing to do is to get rid of A. I think you meant divide by the leading coefficient, but said A by mistake. In this person's formula, A is not the leading coefficient, it is the coefficient of the quadratic term, and B is the coefficient of the linear term. I think that's what you meant anyway.

 

You then say that the next thing to do, is to get rid of B, but I think you meant get rid of A.

 

So let

 

[math] z = x- \frac{A}{3} [/math]

 

Substituting we get:

 

[math] (x- \frac{A}{3})^3 + A(x- \frac{A}{3})^2 + B(x- \frac{A}{3}) + C = 0 [/math]

 

Now:

let b=-A/3

[math] (x +b)^3 = x^3+3x^2b+3xb^2+b^3 [/math]

 

AND

 

[math] (x+b)^3 + A(x+b)^2 + B(x+b) + C = 0 [/math]

 

So that we have:

 

[math] [x^3+3x^2b+3xb^2+b^3]+ A[x^2+b^2+2xb]+B(x+b)+C=0 [/math]

 

This stinks.

[uncool]

Right, yeah, was thinking A*x^3+B*x^2+C*x+D. Well, you can let a calculator do it, or you can finish out the calculation. It takes less time than you might expect.

-Uncool-

[Johnny5]

Right' date=' yeah, was thinking A*x^3+B*x^2+C*x+D. Well, you can let a calculator do it, or you can finish out the calculation. It takes less time than you might expect.

-Uncool-[/quote']

 

I was thinking of a better way to do it this morning.

 

Suppose you already knew one of the three roots, call it r1.

 

So you are given this:

 

[math] a_3Z^3+a_2Z^2+a_1Z+a_0 = 0 [/math]

 

a_3 nonzero, the other coefficients might or might not be zero.

 

So divide everything by the leading coefficient a3, to get:

 

[math] Z^3+AZ^2+BZ+C= 0 [/math]

 

Which is what the original person gave.

 

Now, since you already know one of the three roots, r1, it follows that you can factor the equation above, like so:

 

(Z-r1)(Z^2+DZ+E)=0

 

Right?

 

Now distribute...

 

[math] Z^3+DZ^2+EZ-r_1Z^2-r_1DZ-r_1E = 0 [/math]

 

Now group things in decending powers of Z.

 

[math] Z^3+(D-r_1)Z^2+(E-r_1D)Z-r_1E = 0 [/math]

 

And the given expression was:

 

[math] Z^3+AZ^2+BZ+C= 0 [/math]

 

From which it follows that:

 

[math] A = (D-r_1) [/math]

[math] B = (E-r_1D) [/math]

[math] C = -r_1E [/math]

 

Now assume that you know the quadratic formula.

 

Since you already knew one root, it follows that you can now figure out the remaining two roots... r₂ and r₃ using the quadratic formula.

 

In other words you already know this:

 

(Z-r1)(Z^2+DZ+E)=0

 

So now focus on the following quadratic equation:

 

Z^2+DZ+E=0

 

If you don't already know the quadratic formula, then you can complete the square. But I am presuming that you do know it. Hence the other two roots are:

 

[math] r_2 = \frac{-D+\sqrt{D^2-4E}}{2} [/math]

 

[math] r_3 = \frac{-D-\sqrt{D^2-4E}}{2} [/math]

 

 

So let's try an example problem now.

 

Someone sends you the following cubic equation, at random, and asks you to find all three roots:

 

[math]

3n^3-9n^2+15n-1 =0

[/math]

 

Step 1: Divide both sides by 3, in order to make the leading coefficient of the polynomial equal to 1.

 

[math]

n^3-3n^2+5n-\frac{1}{3} =0

[/math]

 

Hence:

 

[math] A = -3 [/math]

[math] B = 5 [/math]

[math] C = - \frac{1}{3} [/math]

 

So that we can now write the given equation as:

 

 

[math]

n^3+An^2+Bn+C =0

[/math]

 

Let us presume that we already know r1, and can factor it out, like so:

 

[math] (n-r_1)(n^2+Dn+E) = 0 [/math]

 

 

Using the quadratic formula, we must have:

 

[math] r_2 = \frac{-D+\sqrt{D^2-4E}}{2} [/math]

 

[math] r_3 = \frac{-D-\sqrt{D^2-4E}}{2} [/math]

 

 

And the relationships between D,E, to A,B,C are:

 

[math] A = (D-r_1) = -3[/math]

[math] B = (E-r_1D) = 5 [/math]

[math] C = -r_1E = -\frac{1}{3} [/math]

 

So

D = -3+r1

r1D = E-5

[math] E= \frac{1}{3r_1} [/math]

 

So

 

[math] r_1D = \frac{1}{3r_1} -5 [/math]

 

And also:

 

[math] r_1 D = -3r_1 + r_1^2 [/math]

 

From which it follows that:

 

[math] \frac{1}{3r_1} -5 = -3r_1 + r_1^2 [/math]

 

From which it follows that:

 

[math] 1 -5r_1 = -3r_1^2 + r_1^3 [/math]

 

From which it follows that:

 

[math] 0 = -3r_1^2 + r_1^3 -1+5r_1[/math]

 

Or equivalently:

 

[math] r_1^3-3r_1^2+5r_1-1 = 0[/math]

 

Now, we also know that:

 

[math] r_2 = \frac{-D+\sqrt{D^2-4E}}{2} [/math]

 

[math] r_3 = \frac{-D-\sqrt{D^2-4E}}{2} [/math]

 

Hence, we can replace D by -3+r1, to obtain:

 

[math]

r_2 = \frac{-(-3+r_1)+\sqrt{(-3+r_1)^2-4E}}{2}

[/math]

 

and

 

[math]

r_3 = \frac{-(-3+r_1)-\sqrt{(-3+r_1)^2-4E}}{2}

[/math]

 

And we can replace E by [math] \frac{1}{3r_1} [/math] to obtain:

 

[math]

r_2 = \frac{-(-3+r_1)+\sqrt{(-3+r_1)^2- \frac{4}{3r_1}}}{2}

[/math]

 

and

 

[math]

r_3 = \frac{-(-3+r_1)-\sqrt{(-3+r_1)^2-\frac{4}{3r_1}}}{2}

[/math]

 

So, we now have the following three formulas:

 

[math] r_1^3-3r_1^2+5r_1-1 = 0[/math]

 

[math] 2r_2 = (-3-r_1)+\sqrt{(-3+r_1)^2- \frac{4}{3r_1}}}

[/math]

 

[math] 2r_3 = (-3-r_1)-\sqrt{(-3+r_1)^2-\frac{4}{3r_1}}}

[/math]

[uncool]

Correct, I was showing how to find the first root of the equation.

-Uncool-

[Johnny5]

Correct' date=' I was showing how to find the first root of the equation.

-Uncool-[/quote']

 

Will the approach I am using above work?

[Johnny5]

I was thinking of a better way to do it this morning.

 

Suppose you already knew one of the three roots' date=' call it r1.

 

So you are given this:

 

[math'] a_3Z^3+a_2Z^2+a_1Z+a_0 = 0 [/math]

 

a_3 nonzero, the other coefficients might or might not be zero.

 

So divide everything by the leading coefficient a3, to get:

 

[math] Z^3+AZ^2+BZ+C= 0 [/math]

 

Which is what the original person gave.

 

Now, since you already know one of the three roots, r1, it follows that you can factor the equation above, like so:

 

(Z-r1)(Z^2+DZ+E)=0

 

Right?

 

Now distribute...

 

[math] Z^3+DZ^2+EZ-r_1Z^2-r_1DZ-r_1E = 0 [/math]

 

Now group things in decending powers of Z.

 

[math] Z^3+(D-r_1)Z^2+(E-r_1D)Z-r_1E = 0 [/math]

 

And the given expression was:

 

[math] Z^3+AZ^2+BZ+C= 0 [/math]

 

From which it follows that:

 

[math] A = (D-r_1) [/math]

[math] B = (E-r_1D) [/math]

[math] C = -r_1E [/math]

 

Now assume that you know the quadratic formula.

 

Since you already knew one root, it follows that you can now figure out the remaining two roots... r₂ and r₃ using the quadratic formula.

 

In other words you already know this:

 

(Z-r1)(Z^2+DZ+E)=0

 

So now focus on the following quadratic equation:

 

Z^2+DZ+E=0

 

If you don't already know the quadratic formula, then you can complete the square. But I am presuming that you do know it. Hence the other two roots are:

 

[math] r_2 = \frac{-D+\sqrt{D^2-4E}}{2} [/math]

 

[math] r_3 = \frac{-D-\sqrt{D^2-4E}}{2} [/math]

 

 

So let's try an example problem now.

 

Someone sends you the following cubic equation, at random, and asks you to find all three roots:

 

[math]

3n^3-9n^2+15n-1 =0

[/math]

 

Step 1: Divide both sides by 3, in order to make the leading coefficient of the polynomial equal to 1.

 

[math]

n^3-3n^2+5n-\frac{1}{3} =0

[/math]

 

Hence:

 

[math] A = -3 [/math]

[math] B = 5 [/math]

[math] C = - \frac{1}{3} [/math]

 

So that we can now write the given equation as:

 

 

[math]

n^3+An^2+Bn+C =0

[/math]

 

Let us presume that we already know r1, and can factor it out, like so:

 

[math] (n-r_1)(n^2+Dn+E) = 0 [/math]

 

 

Using the quadratic formula, we must have:

 

[math] r_2 = \frac{-D+\sqrt{D^2-4E}}{2} [/math]

 

[math] r_3 = \frac{-D-\sqrt{D^2-4E}}{2} [/math]

 

 

And the relationships between D,E, to A,B,C are:

 

[math] A = (D-r_1) = -3[/math]

[math] B = (E-r_1D) = 5 [/math]

[math] C = -r_1E = -\frac{1}{3} [/math]

 

So

D = -3+r1

r1D = E-5

[math] E= \frac{1}{3r_1} [/math]

 

So

 

[math] r_1D = \frac{1}{3r_1} -5 [/math]

 

And also:

 

[math] r_1 D = -3r_1 + r_1^2 [/math]

 

From which it follows that:

 

[math] \frac{1}{3r_1} -5 = -3r_1 + r_1^2 [/math]

 

From which it follows that:

 

[math] 1 -5r_1 = -3r_1^2 + r_1^3 [/math]

 

From which it follows that:

 

[math] 0 = -3r_1^2 + r_1^3 -1+5r_1[/math]

 

Or equivalently:

 

[math] r_1^3-3r_1^2+5r_1-1 = 0[/math]

 

Now, we also know that:

 

[math] r_2 = \frac{-D+\sqrt{D^2-4E}}{2} [/math]

 

[math] r_3 = \frac{-D-\sqrt{D^2-4E}}{2} [/math]

 

Hence, we can replace D by -3+r1, to obtain:

 

[math]

r_2 = \frac{-(-3+r_1)+\sqrt{(-3+r_1)^2-4E}}{2}

[/math]

 

and

 

[math]

r_3 = \frac{-(-3+r_1)-\sqrt{(-3+r_1)^2-4E}}{2}

[/math]

 

And we can replace E by [math] \frac{1}{3r_1} [/math] to obtain:

 

[math]

r_2 = \frac{-(-3+r_1)+\sqrt{(-3+r_1)^2- \frac{4}{3r_1}}}{2}

[/math]

 

and

 

[math]

r_3 = \frac{-(-3+r_1)-\sqrt{(-3+r_1)^2-\frac{4}{3r_1}}}{2}

[/math]

 

So, we now have the following three formulas:

 

[math] r_1^3-3r_1^2+5r_1-1 = 0[/math]

 

[math] 2r_2 = (-3-r_1)+\sqrt{(-3+r_1)^2- \frac{4}{3r_1}}}

[/math]

 

[math] 2r_3 = (-3-r_1)-\sqrt{(-3+r_1)^2-\frac{4}{3r_1}}}

[/math]

 

The question is, "can we now find the specific numbers r1,r2,r3, from the three equations immediately above?"

 

Suppose that we add two of the three equations together, like so:

 

[math] 2r_2 + 2r_3 = 2 (-3-r_1) [/math]

 

And now divide both sides by two, to obtain:

 

[math] r_2 + r_3 = -3-r_1 [/math]

 

It now follows that:

 

[math] r_1+r_2 + r_3 = -3 [/math]

 

Thus, the sum of the roots corresponds to the product of the leading coefficient and the constant term in the original formula

 

[math]

 

3n^3-9n^2+15n-1 =0

 

[/math]

 

I don't know if there is a theorem for this, but it seems familiar for some reason.

 

Ideally, we want three equations in three unknowns, so that we can then solve the system using matrices.

 

If instead, we subtract 2r3 from 2r2, we obtain:

 

[math] 2r_2-2r_3 = 2\sqrt{(-3+r_1)^2- \frac{4}{3r_1}}} [/math]

 

And now, division of both sides by 2 yields:

 

[math] r_2-r_3 = \sqrt{(-3+r_1)^2- \frac{4}{3r_1}}} [/math]

 

If we now square both sides we obtain and expression which is devoid of the radical sign:

 

[math] (r_2-r_3)^2 = (-3+r_1)^2- \frac{4}{3r_1}} [/math]

 

If we multiply both sides by r1 we obtain:

 

[math] r_1(r_2-r_3)^2 = r_1(-3+r_1)^2- 4 [/math]

 

Now, the RHS is a cubic in r1, whereas the LHS is going to be quadratic in r2,r3. I am going to try to make use of this fact. If it leads nowhwere, then I'm going to give up.

 

[math] r_1(r_2-r_3)^2 = r_1[9+r_1^2-6r_1]-4 [/math]

 

[math] r_1(r_2-r_3)^2 = 9r_1+r_1^3-6r_1^2-4 [/math]

 

From an earlier line of work we know that:

 

[math]

r_1^3-3r_1^2+5r_1-1 = 0

[/math]

 

Solving for r1 cubed gives:

 

[math]

r_1^3 = 3r_1^2-5r_1+1

[/math]

 

Therefore it follows that:

 

[math] r_1(r_2-r_3)^2 = 9r_1+(3r_1^2-5r_1+1 )-6r_1^2-4 [/math]

 

Combining like terms on the RHS leads to:

 

[math] r_1(r_2-r_3)^2 = -3r_1^2+4r_1-3 [/math]

 

Which leads to:

 

[math] -3r_1^2+4r_1-3 - r_1(r_2-r_3)^2 =0 [/math]

 

Therefore:

 

[math] -3r_1^2+(4 - (r_2-r_3)^2)r_1 -3 =0 [/math]

 

The previous equation is quadratic in r1, so in principle, if we knew r2, and r3, then we could solve it for r1 using the quadratic formula.

 

Divide both sides of the equation above by negative three to obtain:

 

[math] r_1^2+ ( \frac{-4 + (r_2-r_3)^2}{3} ) r_1 +1 =0 [/math]

 

 

Now, to lighten up on the reading, define xi such that:

 

[math] \xi \equiv( \frac{-4 + (r_2-r_3)^2}{3} ) [/math]

 

And rewrite the equation above as:

 

[math] r_1^2+ \xi r_1 +1 =0 [/math]

 

The roots of the equation above, by the quadratic formula are:

 

[math] \text{root 1} = \frac{-\xi + \sqrt{\xi^2 - 4}}{2} [/math]

 

And

 

[math] \text{root 2} = \frac{-\xi - \sqrt{\xi^2 - 4}}{2} [/math]

 

 

Now, r1 is a root of the principle equation which was:

 

[math]

 

 

3n^3-9n^2+15n-1 =0

 

 

[/math]

 

And we cannot have two solutions for r1, as the above roots, labeled 'root 1' and 'root 2' suggest. So one of them will lead to an explicit contradiction.

 

Hence root 1=root 2 if and only if

 

[math] \sqrt{\xi^2 - 4} = 0 [/math]

 

Which is true if and only if

 

[math] \xi^2 - 4 = 0 [/math]

 

Which is true if and only if

 

[math] \xi^2 = 4 [/math]

 

Which is true if and only if

 

[math] \xi = 2 [/math]

 

Which is true if and only if

 

 

[math] \frac{-4 + (r_2-r_3)^2}{3} = 2 [/math]

 

Which is true if and only if

 

[math] -4 + (r_2-r_3)^2 = 6 [/math]

 

Which is true if and only if

 

[math] (r_2-r_3)^2 = 10 [/math]

 

Which is true if and only if

 

[math] r_2-r_3 = \sqrt{10} [/math]

 

And this result is new.

 

We now have two equations in three unknowns.

 

[math] \text{Equation 1}: r_1+r_2 + r_3 = -3 [/math]

 

[math] \text{Equation 2}: 0r_1+ r_2-r_3 = \sqrt{10}[/math]

 

Now, we just need one more equation in the same three unknowns, and we can use the techniques of linear algebra, to find the unique solution.

[uncool]

Hmm, can't see the LATEX stuff - it's giving me a "couldn't open URL" error.

-Uncool-

So to reiterate:

To find the first solution:

Take out the x^2 term by substituting y - A/3 for x.

Let this be y^3 + p*y+q = 0

Let y = z - p/3z (I think)

This should form an equation of the form:

z^3 + q + r/z^3 = 0

z^6 + qz^3 + r = 0

Find z^3, take the cube root, and substitute back into y, then into x.

That should be the first solution, and, as you showed, finding the other two is easy.

-Uncool-

[Dave]

Unfortunately it appears that we're having problems with the off-site latex. Should be up soon, hopefully.

[Johnny5]

Hmm' date=' can't see the LATEX stuff - it's giving me a "couldn't open URL" error.

-Uncool-

So to reiterate:

To find the first solution:

Take out the x^2 term by substituting y - A/3 for x.

Let this be y^3 + p*y+q = 0

Let y = z - p/3z (I think)

This should form an equation of the form:

z^3 + q + r/z^3 = 0

z^6 + qz^3 + r = 0

Find z^3, take the cube root, and substitute back into y, then into x.

That should be the first solution, and, as you showed, finding the other two is easy.

-Uncool-[/quote']

 

The only problem I have with that method, is that I won't remember it.

 

Also, I located two tiny errors in my work above, and I was going to correct them today, and then find a third equation in the same three unknowns, and thus effectively bypass Cardan's method altogether.

 

I like the method I've found better, if it works.

 

But since the latex isn't working, I will wait.

 

Regards

[Johnny5]

This was something i was working on awhile back, and wanted to see through to the end, but then Latex went down. Since Latex is finally back up, its been about a month, I'm going to resume this.

 

The whole issue had to do with Cardan's formula.

 

It's been so long that I'm going to begin again fresh.

 

It began in post 8, with this:

 

I was thinking of a better way to do it this morning.

 

Suppose you already knew one of the three roots' date=' call it r1.

 

So you are given this:

 

The following code was used to generate the image you clicked on:

 

[math']

a_3Z^3+a_2Z^2+a_1Z+a_0 = 0

[/math]

 

a_3 nonzero, the other coefficients might or might not be zero.

 

So divide everything by the leading coefficient a3, to get:

 

[math]

Z^3+AZ^2+BZ+C= 0

[/math]

 

Which is what the original person gave.

 

So I am going to pick it up from here.

[uncool]

OK, to remember it, just remember that you want to eliminate the middle terms. First the A term, then the B term.

 

To eliminate the A term: Let [math]Y = Z + A/3[/math]

 

Let this leave you with [math]Y^3 + PY + Q = 0[/math]

 

To eliminate the P term: Let [math]Y = X-P/(3X)[/math]

 

Let this leave you with [math]X^3 + M - N/X^3 = 0[/math]

 

Find X^3 because this is a quadratic equation. Both will give the same final answer.

 

Backsolve to find Z.

 

Use this solution to reduce the cubic to a quadratic.

 

Solve the quadratic for the other two roots.

 

So pretty much all you have to do is remember how to eliminate each middle term.

-Uncool-

[BigMoosie]

I found this equation for ax^3 + bx^2 + cx + d = 0:

 

 

It works for many cubics unless the roots create imaginaries in which case you must handle complex numbers that cancel out to eventually give you a real root. This always gives the root that has to exist, what would be the formula for the other two? Also, is it possible to expand the above formula so that one does not require knowledge of imaginaries and it would be possible to feed into a computer to solve?

[Dave]

Well, it's perfectly okay for a cubic to have two complex roots. Computers can deal with such objects if you're willing to put in the effort to tell the computer how to handle them (or alternatively, use something like Fortran which already has a complex data type).

[Johnny5]

[math]

Z^3+AZ^2+BZ+C= 0

[/math]

 

Presume that you already know one root r1. It follows that you can factor the equation like so:

 

(Z-r1)(Z^2+DZ+E)=0

 

Now, distribute to get:

 

[math]

Z^3+DZ^2+EZ-r_1Z^2-r_1DZ-r_1E = 0

[/math]

 

 

Now, group things in descending powers of Z, to obtain:

[math]

Z^3+(D-r_1)Z^2+(E-r_1D)Z-r_1E = 0

[/math]

 

And the given expression was:

[math]

Z^3+AZ^2+BZ+C= 0

[/math]

 

from which it follows that:

[math]

A = (D-r_1)

[/math]

[math]

B = (E-r_1D)

[/math]

[math]

C = r_1E

[/math]

 

Since you already knew one root, it follows that you can now figure out the remaining two roots... r2 and r3 using the quadratic formula.

 

In other words you already know this:

 

(Z-r1)(Z^2+DZ+E)=0

 

So now focus on the following quadratic equation:

 

Z^2+DZ+E=0

 

Using the quadratic formula, the remaining two roots are given by:

 

[math]

r_2 = \frac{-D+\sqrt{D^2-4E}}{2}

[/math]

 

[math]

r_2 = \frac{-D-\sqrt{D^2-4E}}{2}

[/math]

 

At this point, i wanted to solve an example problem.

 

You are given the following cubic equation, and asked to find all three roots:

 

[math]

3n^3-9n^2+15n-1 =0

[/math]

 

The first thing you do, is make the leading coefficient equal to 1, like so:

 

[math]

n^3-3n^2+5n-\frac{1}{3} =0

[/math]

 

Let

A= -3

B=5

c=-1/3

 

We can now write the equation as:

 

[math]

n^3+An^2+Bn+C =0

[/math]

 

Let us presume that we already know r1, and can factor it out like so:

 

[math]

(n-r_1)(n^2+Dn+E) = 0

[/math]

 

Now, the roots of the quadratic are:

 

[math] r_2 = \frac{-D + \sqrt{D^2-4E}}{2} [/math]

 

[math] r_3 = \frac{-D - \sqrt{D^2-4E}}{2} [/math]

 

And the relationships between D,E and A,B,C are:

 

[math]

A = (D-r_1)= -3

[/math]

[math]

B = (E-r_1D)= 5

[/math]

[math]

C = r_1E = -\frac{1}{3}

[/math]

 

Hence:

 

[math]

D=r_1-3

[/math]

[math]

r_1D= E-5

[/math]

[math]

E= \frac{1}{3r_1}

[/math]

 

Hence:

[math]

r_1D= \frac{1}{3r_1}-5

[/math]

 

From which it follows that:

 

[math]

\frac{1}{3r_1} -5 = -3r_1 + r_1^2

[/math]

 

From which it follows that:

 

[math]

r_1^3-3r_1^2+5r_1-1 = 0

[/math]

 

 

Now, we also know that:

 

[math] r_2 = \frac{-D+\sqrt{D^2-4E}}{2} [/math]

 

[math] r_3 = \frac{-D-\sqrt{D^2-4E}}{2} [/math]

 

Hence, we can replace D by -3+r1, to obtain:

 

[math]

r_2 = \frac{-(-3+r_1)+\sqrt{(-3+r_1)^2-4E}}{2}

[/math]

 

and

 

[math]

r_3 = \frac{-(-3+r_1)-\sqrt{(-3+r_1)^2-4E}}{2}

[/math]

 

And we can replace E by [math] \frac{1}{3r_1} [/math] to obtain:

 

[math]

r_2 = \frac{-(-3+r_1)+\sqrt{(-3+r_1)^2- \frac{4}{3r_1}}}{2}

[/math]

 

and

 

[math]

r_3 = \frac{-(-3+r_1)-\sqrt{(-3+r_1)^2-\frac{4}{3r_1}}}{2}

[/math]

 

So, we now have the following three formulas:

 

[math] r_1^3-3r_1^2+5r_1-1 = 0[/math]

 

[math] 2r_2 = (3-r_1)+\sqrt{(-3+r_1)^2- \frac{4}{3r_1}}}

[/math]

 

[math] 2r_3 = (3-r_1)-\sqrt{(-3+r_1)^2-\frac{4}{3r_1}}}

[/math]

 

The question is, "can we now find the specific numbers r1,r2,r3, from the three equations immediately above?"

 

Suppose that we add two of the three equations together, like so:

 

[math]2r_2 +2r_3 = 2(3-r_1)

[/math]

 

From which it follows that:

 

[math] r_1+r_2+r_3 = 3[/math]

 

If we now subtract 2r3 from 2r2 we obtain:

 

[math]

2r_2-2r_3 = 2\sqrt{(-3+r_1)^2- \frac{4}{3r_1}}}

[/math]

 

Dividing both sides by two, and then squaring we obtain:

 

[math]

(r_2-r_3)^2 = (-3+r_1)^2- \frac{4}{3r_1}}

[/math]

 

 

If we multiply both sides by 3r1 we obtain:

 

[math] 3r_1(r_2-r_3)^2 = 3r_1(-3+r_1)^2- 4 [/math]

 

Now, the RHS is a cubic in r1, whereas the LHS is going to be quadratic in r2,r3.

 

[math] 3r_1(r_2-r_3)^2 = 3r_1[9+r_1^2-6r_1]-4 [/math]

 

[math] 3r_1(r_2-r_3)^2 = 27r_1+3r_1^3-18r_1^2-4 [/math]

 

From an earlier line of work we know that:

 

[math]

r_1^3-3r_1^2+5r_1-1 = 0

[/math]

 

Solving for r1 cubed gives:

 

[math]

r_1^3 = 3r_1^2-5r_1+1

[/math]

 

Therefore it follows that:

 

[math]

3r_1(r_2-r_3)^2 = 27r_1+3(3r_1^2-5r_1+1 )-18r_1^2-4

[/math]

 

From which it follows that:

 

[math] 3r_1(r_2-r_3)^2 = 27r_1+9r_1^2-15r_1+3 -18r_1^2-4 [/math]

 

Combining like terms on the RHS leads to:

 

[math] 3r_1(r_2-r_3)^2 = -9r_1^2+12r_1-1 [/math]

 

Which leads to:

 

[math] 9r_1^2-12r_1+1 +3r_1(r_2-r_3)^2 =0 [/math]

 

Therefore:

 

[math] 9r_1^2+r_1(-12+3(r_2-r_3)^2)+1=0 [/math]

 

The previous equation is quadratic in r1, so in principle, if we knew r2, and r3, then we could solve it for r1 using the quadratic formula.

 

Divide both sides of the equation above by negative three to obtain:

 

[math] r_1^2+ ( \frac{-4 + (r_2-r_3)^2}{3} ) r_1 +1 =0 [/math]

 

 

Now, to lighten up on the reading, define xi such that:

 

[math] \xi \equiv( \frac{-4 + (r_2-r_3)^2}{3} ) [/math]

 

And rewrite the equation above as:

 

[math] r_1^2+ \xi r_1 +1 =0 [/math]

 

The roots of the equation above, by the quadratic formula are:

 

[math] \text{root 1} = \frac{-\xi + \sqrt{\xi^2 - 4}}{2} [/math]

 

And

 

[math] \text{root 2} = \frac{-\xi - \sqrt{\xi^2 - 4}}{2} [/math]

 

 

Now, r1 is a root of the principle equation which was:

 

[math]

 

 

3n^3-9n^2+15n-1 =0

 

 

[/math]

 

And we cannot have two solutions for r1, as the above roots, labeled 'root 1' and 'root 2' suggest. So one of them will lead to an explicit contradiction.

 

Hence root 1=root 2 if and only if

 

[math] \sqrt{\xi^2 - 4} = 0 [/math]

 

Which is true if and only if

 

[math] \xi^2 - 4 = 0 [/math]

 

Which is true if and only if

 

[math] \xi^2 = 4 [/math]

 

Which is true if and only if

 

[math] \xi = 2 [/math]

 

Which is true if and only if

 

 

[math] \frac{-4 + (r_2-r_3)^2}{3} = 2 [/math]

 

Which is true if and only if

 

[math] -4 + (r_2-r_3)^2 = 6 [/math]

 

Which is true if and only if

 

[math] (r_2-r_3)^2 = 10 [/math]

 

Which is true if and only if

 

[math] r_2-r_3 = \sqrt{10} [/math]

 

And this result is new.

 

We now have two equations in three unknowns.

 

[math] \text{Equation 1}: r_1+r_2 + r_3 = -3 [/math]

 

[math] \text{Equation 2}: 0r_1+ r_2-r_3 = \sqrt{10}[/math]

 

Now, we just need one more equation in the same three unknowns, and we can use the techniques of linear algebra, to find the unique solution.

[uncool]

From where are you getting r1, Johnny?

Or do you mean to try to find r1 from the coefficients without doing what I did?

-Uncool-

[Johnny5]

From where are you getting r1' date=' Johnny?

Or do you mean to try to find r1 from the coefficients without doing what I did?

-Uncool-[/quote']

 

I am trying to find a simpler way to solve a cubic, without doing what you did.

[Johnny5]

I've been trying to fix an error that propagated through my original attempt, and unfortunately you only have six hours to edit any given post, but the information has to be kept in one place to follow the argument. What I am trying to do, is come up with a method of solving a cubic equation which is better than Cardan's formula.

 

Here is what I have so far:

[math]

Z^3+AZ^2+BZ+C= 0

[/math]

 

Presume that you already know one root r1. It follows that you can factor the equation like so:

 

(Z-r1)(Z^2+DZ+E)=0

 

Now' date=' distribute to get:

 

[math']

Z^3+DZ^2+EZ-r_1Z^2-r_1DZ-r_1E = 0

[/math]

 

 

Now, group things in descending powers of Z, to obtain:

[math]

Z^3+(D-r_1)Z^2+(E-r_1D)Z-r_1E = 0

[/math]

 

And the given expression was:

[math]

Z^3+AZ^2+BZ+C= 0

[/math]

 

from which it follows that:

[math]

A = (D-r_1)

[/math]

[math]

B = (E-r_1D)

[/math]

[math]

C = r_1E

[/math]

 

Since you already knew one root, it follows that you can now figure out the remaining two roots... r2 and r3 using the quadratic formula.

 

In other words you already know this:

 

(Z-r1)(Z^2+DZ+E)=0

 

So now focus on the following quadratic equation:

 

Z^2+DZ+E=0

 

Using the quadratic formula, the remaining two roots are given by:

 

[math]

r_2 = \frac{-D+\sqrt{D^2-4E}}{2}

[/math]

 

[math]

r_2 = \frac{-D-\sqrt{D^2-4E}}{2}

[/math]

 

At this point, i wanted to solve an example problem.

 

You are given the following cubic equation, and asked to find all three roots:

 

[math]

3n^3-9n^2+15n-1 =0

[/math]

 

The first thing you do, is make the leading coefficient equal to 1, like so:

 

[math]

n^3-3n^2+5n-\frac{1}{3} =0

[/math]

 

Let

A= -3

B=5

c=-1/3

 

We can now write the equation as:

 

[math]

n^3+An^2+Bn+C =0

[/math]

 

Let us presume that we already know r1, and can factor it out like so:

 

[math]

(n-r_1)(n^2+Dn+E) = 0

[/math]

 

Now, the roots of the quadratic are:

 

[math] r_2 = \frac{-D + \sqrt{D^2-4E}}{2} [/math]

 

[math] r_3 = \frac{-D - \sqrt{D^2-4E}}{2} [/math]

 

And the relationships between D,E and A,B,C are:

 

[math]

A = (D-r_1)= -3

[/math]

[math]

B = (E-r_1D)= 5

[/math]

[math]

C = r_1E = -\frac{1}{3}

[/math]

 

Hence:

 

[math]

D=r_1-3

[/math]

[math]

r_1D= E-5

[/math]

[math]

E= \frac{1}{3r_1}

[/math]

 

Hence:

[math]

r_1D= \frac{1}{3r_1}-5

[/math]

 

From which it follows that:

 

[math]

\frac{1}{3r_1} -5 = -3r_1 + r_1^2

[/math]

 

From which it follows that:

 

[math]

r_1^3-3r_1^2+5r_1-1 = 0

[/math]

 

 

Now, we also know that:

 

[math] r_2 = \frac{-D+\sqrt{D^2-4E}}{2} [/math]

 

[math] r_3 = \frac{-D-\sqrt{D^2-4E}}{2} [/math]

 

Hence, we can replace D by -3+r1, to obtain:

 

[math]

r_2 = \frac{-(-3+r_1)+\sqrt{(-3+r_1)^2-4E}}{2}

[/math]

 

and

 

[math]

r_3 = \frac{-(-3+r_1)-\sqrt{(-3+r_1)^2-4E}}{2}

[/math]

 

And we can replace E by [math] \frac{1}{3r_1} [/math] to obtain:

 

[math]

r_2 = \frac{-(-3+r_1)+\sqrt{(-3+r_1)^2- \frac{4}{3r_1}}}{2}

[/math]

 

and

 

[math]

r_3 = \frac{-(-3+r_1)-\sqrt{(-3+r_1)^2-\frac{4}{3r_1}}}{2}

[/math]

 

So, we now have the following three formulas:

 

[math] r_1^3-3r_1^2+5r_1-1 = 0[/math]

 

[math] 2r_2 = (3-r_1)+\sqrt{(-3+r_1)^2- \frac{4}{3r_1}}}

[/math]

 

[math] 2r_3 = (3-r_1)-\sqrt{(-3+r_1)^2-\frac{4}{3r_1}}}

[/math]

 

The question is, "can we now find the specific numbers r1,r2,r3, from the three equations immediately above?"

 

Suppose that we add two of the three equations together, like so:

 

[math]2r_2 +2r_3 = 2(3-r_1)

[/math]

 

From which it follows that:

 

[math] r_1+r_2+r_3 = 3[/math]

 

If we now subtract 2r3 from 2r2 we obtain:

 

[math]

2r_2-2r_3 = 2\sqrt{(-3+r_1)^2- \frac{4}{3r_1}}}

[/math]

 

Dividing both sides by two, and then squaring we obtain:

 

[math]

(r_2-r_3)^2 = (-3+r_1)^2- \frac{4}{3r_1}}

[/math]

 

 

If we multiply both sides by 3r1 we obtain:

 

[math] 3r_1(r_2-r_3)^2 = 3r_1(-3+r_1)^2- 4 [/math]

 

Now, the RHS is a cubic in r1, whereas the LHS is going to be quadratic in r2,r3.

 

[math] 3r_1(r_2-r_3)^2 = 3r_1[9+r_1^2-6r_1]-4 [/math]

 

[math] 3r_1(r_2-r_3)^2 = 27r_1+3r_1^3-18r_1^2-4 [/math]

 

From an earlier line of work we know that:

 

[math]

r_1^3-3r_1^2+5r_1-1 = 0

[/math]

 

Solving for r1 cubed gives:

 

[math]

r_1^3 = 3r_1^2-5r_1+1

[/math]

 

Therefore it follows that:

 

[math]

3r_1(r_2-r_3)^2 = 27r_1+3(3r_1^2-5r_1+1 )-18r_1^2-4

[/math]

 

From which it follows that:

 

[math] 3r_1(r_2-r_3)^2 = 27r_1+9r_1^2-15r_1+3 -18r_1^2-4 [/math]

 

Combining like terms on the RHS leads to:

 

[math] 3r_1(r_2-r_3)^2 = -9r_1^2+12r_1-1 [/math]

 

Which leads to:

 

[math] 9r_1^2-12r_1+1 +3r_1(r_2-r_3)^2 =0 [/math]

 

Therefore:

 

[math] 9r_1^2+r_1(-12+3(r_2-r_3)^2)+1=0 [/math]

 

The previous equation is quadratic in r1, so in principle, if we knew r2, and r3, then we could solve it for r1 using the quadratic formula.

 

Now, this is where i left off yesterday. So I am going to pick up the argument from here today, and try to finish it off by the end of the day.

 

To quickly recapitulate, suppose you are given the following random cubic equation:

 

[math]

3n^3-9n^2+15n-1 =0

[/math]

 

You want to find three roots, r1,r1,r3, such that:

 

[math] (n-r_1)(n-r_2)(n-r_3)= 0 [/math]

 

Taking the steps outlined above, you discover the following:

 

[math] r_1+r_2+r_3 = 3[/math]

 

and

 

[math] 9r_1^2+r_1(-12+3(r_2-r_3)^2)+1=0 [/math]

 

Now, we started off by assuming that we knew r1, but didn't know r2 or r3. Now, lets reverse this. Suppose we don't know r1, but we do know r2, and r3. We could now figure out r1, from the formula above.

 

 

Divide both sides of the equation above by 9, in order to obtain:

 

[math] r_1^2+r_1 \frac{(-12+3(r_2-r_3)^2)}{9}+ \frac{1}{9} =0 [/math]

 

 

In order to lighten up on the reading, define xi such that:

 

[math] \xi \equiv \frac{(-12+3(r_2-r_3)^2)}{9} [/math]

 

Thus, we can rewrite the equation as;

 

[math] r_1^2 + \xi r_1 + \frac{1}{9} = 0 [/math]

 

I think i've finally got a handle on the arithmetical mistake i made in earlier posts. Previously I had:

 

[math] r_1^2 + \xi r_1 + 1 = 0 [/math]

 

But this is wrong. What I should have had was:

 

[math] r_1^2 + \xi r_1 + \frac{1}{9} = 0 [/math]

 

Ok so... using the quadratic formula, the roots of the equation above are given by:

 

[math] \text{root 1} \frac{-\xi + \sqrt{\xi^2 - \frac{4}{9}}}{2} [/math]

 

[math]

\text{root 2} = \frac{-\xi - \sqrt{\xi^2 - \frac{4}{9}}}{2}

[/math]

 

Now, r1 is a root of the principle equation which was:

 

[math]

3n^3-9n^2+15n-1 =0

[/math]

 

And we cannot have two solutions for r1, as the above roots, labeled 'root 1' and 'root 2' suggest.

 

root 1 = root 2 if and only if

 

[math] \frac{\sqrt{\xi^2 - \frac{4}{9}}}{2} =0 [/math]

 

Which is true if and only if

 

[math] \sqrt{\xi^2 - \frac{4}{9}}} =0 [/math]

 

Which is true if and only if

 

[math] \xi^2 - \frac{4}{9}}} =0 [/math]

 

Which is true if and only if

 

[math] \xi^2 = \frac{4}{9}}} [/math]

 

Now, there are two possible solutions to the equation above, they are:

 

[math] \xi_1 = \frac{2}{3} [/math]

[math] \xi_2 = -\frac{2}{3} [/math]

 

Recal the definition of x

 

[math]

\xi \equiv \frac{(-12+3(r_2-r_3)^2)}{9}

[/math]

 

Hence

 

[math] \frac{2}{3} = \frac{(-12+3(r_2-r_3)^2)}{9} [/math]

 

or

 

[math] -\frac{2}{3} = \frac{(-12+3(r_2-r_3)^2)}{9} [/math]

 

 

Lets just deal with x₁ for now.

 

So we have:

 

[math] \frac{2}{3} = \frac{(-12+3(r_2-r_3)^2)}{9} [/math]

 

which leads to

 

[math] 6 = -12+3(r_2-r_3)^2 [/math]

 

which leads to

 

[math] 18 = 3(r_2-r_3)^2 [/math]

 

which leads to

 

[math] 6 = (r_2-r_3)^2 [/math]

 

Which leads to:

 

 

[math] r_2-r_3 = \pm \sqrt{6} [/math]

 

That is:

 

 

[math] r_2-r_3 = \sqrt{6} [/math]

 

or

 

[math] r_2-r_3 = -\sqrt{6} [/math]

 

Now, lets consider x₂ for now. We have:

 

[math] -\frac{2}{3} = \frac{(-12+3(r_2-r_3)^2)}{9} [/math]

 

This leads to:

 

[math] -6 = -12+3(r_2-r_3)^2 [/math]

 

Which leads to:

 

[math] 6 = 3(r_2-r_3)^2 [/math]

 

Which leads to:

 

[math] 2 = (r_2-r_3)^2 [/math]

 

Which leads to:

 

[math] r_2-r_3 = \pm \sqrt{2}[/math]

 

That is:

 

[math] r_2-r_3 = \sqrt{2}[/math]

 

or

 

[math] r_2-r_3 = -\sqrt{2}[/math]

 

Now, if the two roots found by using the quadratic formula are necessarily complex, then it is meaningless to speak of one root being greater than the other.

 

On the other hand, if they are both real, than one of them is greater than the other.

 

Assume it is the case that r2 > r3.

 

Thus, r2-r3 > 0

 

Therefore:

[math]

r_2-r_3 = \sqrt{6}

[/math]

and

[math]

r_2-r_3 = \sqrt{2}

[/math]

 

Thus, we have two equations in two unknowns, so we can solve for r2,r3.

 

From the second equation we have:

 

[math]

r_3 = r_2-\sqrt{2}

[/math]

 

Substituting that into the first equation we have:

 

[math]

r_2-(r_2-\sqrt{2}) = \sqrt{6}

[/math]

 

From which it follows that:

 

[math] \sqrt{2} = \sqrt{6} [/math]

 

Which is false. Hence, it is not the case that r2>r3.

 

Now, assume that r3>r2, hence r3-r2>0, hence r2-r3 is negative.

 

Therefore:

 

[math]

r_2-r_3 = -\sqrt{6}

[/math]

 

and

 

[math]

r_2-r_3 = -\sqrt{2}

[/math]

 

Again, we arrive at an impossibility, hence the two roots r2,r3 are complex.

 

In earlier lines of work we arrived at the following two equations:

 

[math]

2r_2 = (3-r_1)+\sqrt{(-3+r_1)^2- \frac{4}{3r_1}}}

[/math]

[math]

2r_3 = (3-r_1)-\sqrt{(-3+r_1)^2- \frac{4}{3r_1}}}

[/math]

 

We added them together, subtracted them from one another, now lets multiply them together, to see what we get:

 

[math] 4 r_2 r_3 = (3-r_1)^2 - [(-3+r_1)^2- \frac{4}{3r_1}] [/math]

 

Which leads to:

 

[math] 4 r_2 r_3 = (3-r_1)^2 - (3-r_1)^2+ \frac{4}{3r_1} [/math]

 

Which leads to:

 

[math] 4 r_2 r_3 = \frac{4}{3r_1} [/math]

 

Which leads to:

 

[math] r_2 r_3 = \frac{1}{3r_1} [/math]

 

Which leads to:

 

[math] r_1r_2 r_3 = \frac{1}{3} [/math]

 

Let's treat r2,r3 as complex numbers. They would have shown up from using the quadratic formula, so that we would have:

 

[math] r_2 = a + bi [/math]

[math] r_3 = a - bi [/math]

 

Now, complex roots would show up in pairs, hence r1 would have to be a real number.

 

Thus, we would have to have this:

 

[math] r_1r_2 r_3 = \frac{1}{3} = r_1(a + bi)(a - bi) [/math]

 

From which it follows that:

 

[math] r_1= \frac{1}{3(a + bi)(a - bi)} [/math]

 

From which it follows that:

 

[math] r_1= \frac{1}{3(a^2 + b^2)} [/math]

 

Now, we also know that

 

[math]

r_1+r_2+r_3 = 3

[/math]

 

hence we must have:

 

[math]

\frac{1}{3(a^2 + b^2)}+(a+bi)+(a-bi) = 3

[/math]

 

From which it follows that:

 

[math]

\frac{1}{3(a^2 + b^2)}+2a = 3

[/math]

 

From which it follows that:

 

[math]

\frac{1}{3(a^2 + b^2)} = 3-2a

[/math]

 

From which it follows that:

 

[math]

3(a^2 + b^2) = \frac{1}{3-2a}

[/math]

 

From which it follows that:

 

[math]

3(a^2 + b^2) (3-2a) = 1

[/math]

[Johnny5]

I think assuming the two roots are complex, at this stage, is way too early. I think I would have done better if i would have used r1r2r3=1/3, so let me try that.

 

You are given the following cubic equation, at random, and asked to find all three roots, be they strictly real, or complex.

 

[math]

3n^3-9n^2+15n-1 =0

[/math]

 

The first thing you do, is make the leading coefficient 1.

 

[math]

n^3-3n^2+5n-\frac{1}{3} =0

[/math]

 

Now, assume that you know r1, and that r1 is strictly real. Thus, you can factor the LHS of equation above, like so:

 

[math] (n-r_1)(n^2+Xn+Y) = 0 [/math]

 

Now, the roots of the quadratic are:

 

[math] r_2 = \frac{-X+\sqrt{x^2-4Y}}{2} [/math]

[math] r_3 = \frac{-X-\sqrt{x^2-4Y}}{2} [/math]

 

If they are complex, they are a complex conjugate pair. Otherwise, all three roots are strictly real.

 

Now, distributing we have:

 

[math] (n-r_1)(n^2+Xn+Y) = n^3+Xn^2+Yn-r_1n^2-Xr_1n-Yr_1 = 0 [/math]

 

Writing terms in descending powers of n, we have:

 

[math] n^3 + (X-r_1)n^2 + (Y-Xr_1)n - Yr_1 = 0 [/math]

 

And we had:

 

[math]

n^3-3n^2+5n-\frac{1}{3} =0

[/math]

 

From which we can rapidly see that:

 

[math] X-r_1 = -3 [/math]

[math] Y-Xr_1 = 5 [/math]

[math] - Yr_1 = -\frac{1}{3} [/math]

 

From the first equation we can see that:

 

[math] X= r_1 -3 [/math]

 

Thus, we can write root 2, and root 3 as follows:

 

 

[math] r_2 = \frac{-(r_1 -3 )+\sqrt{(r_1 -3 )^2-4Y}}{2} [/math]

[math] r_3 = \frac{-(r_1 -3 )-\sqrt{(r_1 -3)^2-4Y}}{2} [/math]

 

From the third equation we can see that:

 

[math] Y = \frac{1}{3r_1} [/math]

 

Thus:

 

[math] 2r_2 = -(r_1 -3 )+\sqrt{(r_1 -3 )^2-\frac{4}{3r_1}} [/math]

[math] 2r_3 = -(r_1 -3 )-\sqrt{(r_1 -3)^2-\frac{4}{3r_1}} [/math]

 

Adding them together we find that:

 

[math] 2r_2 +2r_3 = -2(r_1 -3 ) [/math]

 

From which it follows that:

[math] r_2 +r_3 = -(r_1 -3 ) [/math]

[math] r_2 +r_3 = -r_1 +3 [/math]

[math] r_1 + r_2 + r_3 = 3 [/math]

 

Subtracting one from the other we have:

[math] 2r_2 -2r_3= 2\sqrt{(r_1 -3 )^2-\frac{4}{3r_1}} [/math]

 

From which it follows that:

[math] r_2 -r_3= \sqrt{(r_1 -3 )^2-\frac{4}{3r_1}} [/math]

 

Squaring both sides we have:

 

[math] (r_2 -r_3)^2= (r_1 -3 )^2-\frac{4}{3r_1} [/math]

 

Lastly, let us multiply them together. Thus:

 

[math] 2r_2 = -(r_1 -3 )+\sqrt{(r_1 -3 )^2-\frac{4}{3r_1}} [/math]

[math] 2r_3 = -(r_1 -3 )-\sqrt{(r_1 -3)^2-\frac{4}{3r_1}} [/math]

 

 

[math] 4r_2r_3 = -\frac{4}{3r_1} [/math]

 

From which it follows that:

 

[math] r_2r_3 = -\frac{1}{3r_1} [/math]

 

From which it follows that:

 

[math] r_1r_2r_3 = -\frac{1}{3} [/math]

 

Now, earlier we found that:

 

[math]

(r_2 -r_3)^2= (r_1 -3 )^2-\frac{4}{3r_1}

[/math]

 

Multiplying both sides by r_1 leads to:

 

[math]

r_1(r_2 -r_3)^2= r_1(r_1 -3 )^2-\frac{4}{3}

[/math]

 

Notice that the RHS is cubic in r1, and the LHS is quadratic in r2,r3.

 

Recall that:

 

[math]

n^3-3n^2+5n-\frac{1}{3} =0

[/math]

 

And the statement above is true, when n is a root, hence:

 

[math]

r_1^3-3r_1^2+5r_1-\frac{1}{3} =0

[/math]

 

From which we rapidly find that:

 

[math]

r_1^3 = 3r_1^2-5r_1+\frac{1}{3}

[/math]

 

And we had:

 

[math]

r_1(r_2 -r_3)^2= r_1(r_1 -3 )^2-\frac{4}{3}

[/math]

 

From which it follows that:

 

[math]

r_1(r_2 -r_3)^2= r_1(r_1^2-6r_1+9)-\frac{4}{3}

[/math]

 

From which it follows that:

 

[math]

r_1(r_2 -r_3)^2= r_1^3-6r_1^2+9r_1-\frac{4}{3}

[/math]

 

Therefore, we have:

 

[math]

r_1(r_2 -r_3)^2= (3r_1^2-5r_1+\frac{1}{3})-6r_1^2+9r_1-\frac{4}{3}

[/math]

 

From which it follows that:

 

[math]r_1(r_2 -r_3)^2 = -3r_1^2 +4r_1 -1[/math]

 

From which it follows that:

 

[math] r_1(r_2 -r_3)^2 +3r_1^2 -4r_1 +1 = 0[/math]

 

Which we can write as:

 

[math] 3r_1^2 + [(r_2 -r_3)^2 -4] r_1 +1 = 0[/math]

 

Define xi such that

 

[math] \xi \equiv [(r_2 -r_3)^2 -4] [/math]

 

So we have:

 

[math] 3r_1^2 + \xi r_1 +1 = 0[/math]

 

The roots of which are:

 

[math] \text{root 1} = \frac{-\xi+\sqrt{\xi^2-12}}{6} [/math]

[math] \text{root 2} = \frac{-\xi-\sqrt{\xi^2-12}}{6} [/math]

 

If we stipulate that they must be equal, we must have:

 

[math] \sqrt{\xi^2-12} = 0 [/math]

 

From which it follows that:

 

[math] \xi^2=12 [/math]

 

From which it follows that:

 

[math] \xi = \pm \sqrt{12} [/math]

 

Therefore:

 

[math] [(r_2 -r_3)^2 -4] = \pm \sqrt{12} [/math]

 

From which it follows that:

 

[math] (r_2 -r_3)^2 = 4 \pm 2\sqrt{3} [/math]

 

Assume that:

 

[math] (r_2 -r_3)^2 = 4 + 2\sqrt{3} [/math]

 

Thus, it follows that:

 

[math] r_2 -r_3 = \sqrt{4 + 2\sqrt{3}} [/math]

 

 

 

Now, earlier we found that:

 

[math]

(r_2 -r_3)^2= (r_1 -3 )^2-\frac{4}{3r_1}

[/math]

 

And we now know that:

 

[math] 4r_2r_3 = -\frac{4}{3r_1} [/math]

 

Hence, it follows that:

 

[math]

(r_2 -r_3)^2 = (r_1 -3 )^2+4r_2r_3

[/math]

 

Thus, it follows that:

 

[math] 4 + 2\sqrt{3} = (r_1 -3 )^2+4r_2r_3 [/math]

 

Now, recall that

[math]

r_1 + r_2 + r_3 = 3

[/math]

 

Therefore:

 

[math]

r_1= 3 -r_2 -r_3

[/math]

 

Hence:

 

[math] 4 + 2\sqrt{3} = ((3 -r_2 -r_3) -3 )^2+4r_2r_3 [/math]

 

To clean things up, define

 

[math] p \equiv 4 + 2\sqrt{3} [/math]

So

[math] p = ((3 -r_2 -r_3) -3 )^2+4r_2r_3 [/math]

 

And expanding the squared term on the RHS, we have:

 

[math] p = (3 -r_2 -r_3)^2+9-6(3 -r_2 -r_3) +4r_2r_3 [/math]

 

Which leads to:

 

[math] p = 9 +(r_2+r_3)^2-6(r_2+r_3) + 9 -18 +6r_2 +6r_3 + 4r_2r_3 [/math]

 

Which simplifies to:

 

[math] p = 9 +(r_2+r_3)^2 + 9 -18 + 4r_2r_3 [/math]

 

Which simplifies to:

 

[math] p = (r_2+r_3)^2 + 4r_2r_3 [/math]

 

Which leads to:

 

[math] p = r_2^2+r_3^2+2r_2r_3 + 4r_2r_3 [/math]

 

Thus we have:

 

[math] r_2^2 +(6r_3)r_2 + (r_3^2- p) = 0 [/math]

 

Using QF we have:

 

[math] r_2 = \frac{-6r_3 \pm \sqrt{36r_3^2 -4(r_3^2- p)}}{2} [/math]

 

Stipulating that r_2 be unique, we must have:

 

[math] \sqrt{36r_3^2 -4(r_3^2- p)} = 0 [/math]

 

From which it follows that:

 

[math] 36r_3^2 -4(r_3^2- p) = 0 [/math]

 

From which it follows that:

 

 

[math] 36r_3^2 -4r_3^2+4p = 0 [/math]

 

From which it follows that:

 

[math] 32r_3^2 +4p = 0 [/math]

 

From which it follows that:

 

[math] r_3^2 = -\frac{4p}{32} [/math]

 

From which it follows that:

 

[math] r_3 = \pm i \sqrt{\frac{4p}{32}} [/math]

 

Now, assume that:

 

[math] r_3 = i \sqrt{\frac{4p}{32}} [/math]

 

Recall that

 

[math]

r_2 = \frac{-6r_3 \pm \sqrt{36r_3^2 -4(r_3^2- p)}}{2}

[/math]

 

And the quantity inside the square root was set to zero, so that:

 

[math]

r_2 = \frac{-6r_3}{2}

[/math]

 

Hence:

 

[math]

r_2 = -3i \sqrt{\frac{4p}{32}}

[/math]

 

Now, recall that:

 

[math]

r_1 + r_2 + r_3 = 3

[/math]

 

Hence:

 

[math]

r_1 + (-3i \sqrt{\frac{4p}{32}} )+(i \sqrt{\frac{4p}{32}}) = 3 [/math]

 

From which it follows that:

 

[math] r_1 + -2i \sqrt{\frac{4p}{32}} = 3 [/math]

 

From which it follows that:

 

[math] r_1 = 3 + 2i \sqrt{\frac{4p}{32}} [/math]

 

So that r1 is necessarily complex, contrary to the stipulation that it be real. Thus, at least one of the assumptions that was made, is false.

 

Let me leave this post here, since it went fairly well. We got the square root of a square root, which is similar to a formula posted earlier in the thread.

[Johnny5]

As a wild guess to save the work done in the previous post, let me write what was obtained for r1:

 

[math]

r_1 = 3 + 2i \sqrt{\frac{4p}{32}}

[/math]

 

Since i know that complex roots come in complex conjugate pairs, lets assume that the root above is correct, and that a second root is given by:

 

[math]

r_2 = 3 - 2i \sqrt{\frac{4p}{32}}

[/math]

 

Also recall that:

 

[math]

p \equiv 4 + 2\sqrt{3}

[/math]

 

and that

 

[math]

r_1 + r_2 + r_3 = 3

[/math]

 

Hence:

 

[math] 6 + r_3 = 3 [/math]

 

From which it follows that:

 

[math] r_3 = -3 [/math]

 

Which is false by inspection.

 

The principle equation was:

 

[math]

n^3-3n^2+5n-\frac{1}{3} =0

[/math]

 

And substituting we have:

 

[math]

(-3)^3-3(-3)^2+5(-3)-\frac{1}{3} =0

[/math]

 

Which is false.

 

The basic idea isn't wrong, but that idea involved not making any assumptions at all, during the process of finding the roots. Even one assumption injects uncertainty into the process. I was hoping to finish tonight, but it's not going to happen.

 

Uncool's method is probably what ultimately gets found anyway.

 

You find the one root r1 first, and then its a routine matter to find the other two.

 

I was hoping to generate three equations in three unknowns, and then be able so say, now you can solve using matrices, but that still hasn't happened yet.