System of Equations Problem

[CanadaAotS]

so this is the problem:

 

[math]d + e + f = A[/math]

[math]d*e + d*f + e*f = B[/math]

[math]d*e*f = C[/math]

 

Where your trying to isolate d, e and f in terms of A, B and C.

 

Is it impossible? I haven't been able to isolate anything...

 

BTW I got this by figuring out what the factors of

[math]x^3 + Ax^2 + Bx + C[/math]

 

would be:

 

[math](x + d)(x + e)(x + f)[/math]

 

and the system of equations was how I'd figure out what the d, e and f's equaled.

 

anyways, if it is possible to do, please help, this isn't homework or anything just trying to figure out this...

[Dave]

From the second equation, you must have [imath]d(e+f) + ef = B[/imath]. Then use the first equation and send equation to get [imath]d(A-d) + \frac{C}{d} = D[/imath].

 

This isn't a simple equation. In fact it's a cubic, and you'll have a hard time finding the roots from the looks of things

 

If you're interested in solving cubics, you'd be better off looking at Cardano's method (just google for it).

[cosine]

so this is the problem:

 

[math]d + e + f = A[/math]

[math]d*e + d*f + e*f = B[/math]

[math]d*e*f = C[/math]

 

Where your trying to isolate d' date=' e and f in terms of A, B and C.

 

Is it impossible? I haven't been able to isolate anything...

 

BTW I got this by figuring out what the factors of

[math']x^3 + Ax^2 + Bx + C[/math]

 

would be:

 

[math](x + d)(x + e)(x + f)[/math]

 

and the system of equations was how I'd figure out what the d, e and f's equaled.

 

anyways, if it is possible to do, please help, this isn't homework or anything just trying to figure out this...

Don't do it! Haha trust me I spent more time than I should have trying to solve that system. Its related to the roots of a cubic. Whatever answer you get like Dave said will be a cubic, because any of the roots of the cubic will satisfy any of those variable properties, because the conditions you've placed on them are all commutative, so you can switch them around as much as you want.

[CanadaAotS]

This is what I found out from it though...

 

d + e + f = A

de + df + ef = B

def = C

 

if d = C / ef

and d = A - e - f

 

and I then put them into equation 2 I get:

(C / ef)(e) + (A - e - f)(f) + ef = B

C / f + Af - ef - f^2 + ef = B

-f^2 + Af - B + C/f = 0 multiply by f

-f^3 + Af^2 - Bf + C = 0 This is almost the EXACT equation I use to get the system of equations in the first place!!! (x^3 + Ax^2 + Bx + C = 0)

 

Your right cosine, this is pretty screwy system lol

[Dave]

Yes, unfortunately you get a circular argument in this case. There are different ways of approaching the problem, though