CanadaAotS Posted November 21, 2005 Share Posted November 21, 2005 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... Link to comment Share on other sites More sharing options...
Dave Posted November 21, 2005 Share Posted November 21, 2005 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). Link to comment Share on other sites More sharing options...
cosine Posted November 21, 2005 Share Posted November 21, 2005 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. Link to comment Share on other sites More sharing options...
CanadaAotS Posted November 22, 2005 Author Share Posted November 22, 2005 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 Link to comment Share on other sites More sharing options...
Dave Posted November 22, 2005 Share Posted November 22, 2005 Yes, unfortunately you get a circular argument in this case. There are different ways of approaching the problem, though Link to comment Share on other sites More sharing options...
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