Shaun Jones
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I am trying to convert this equation into a fourth order polynomial (aTtop^4 + bTtop^3 + cTtop^2 + dTtop + e = 0).
I think I multiply each section by h/k?
Any help would be appreciated!
Converting an equation to a standard 4th order polynomial.
- Thread starter Shaun Jones
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I am trying to convert this equation into a fourth order polynomial (aTtop^4 + bTtop^3 + cTtop^2 + dTtop + e = 0).
I think I multiply each section by h/k?
Any help would be appreciated!
HallsofIvy
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- Jan 27, 2012
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With yocar04's suggestion, multiplying by h/k, we have
[math]-(h/k)(1- \alpha)F_{SW}- (h/k)F_{other}+ 0.95\cdot 5.67\cdot 10^{-8}(h/k) T_{top}^4= -T_{top}+ T_{bot}[/math].
That can be written
[math]0.95\cdot 5.67\cdot 10^{-8}(h/k) T_{top}^4+ T_{top}+ -(h/k)(1- \alpha)F_{SW}- (h/k)F_{other}- T_{bot}= 0[/math].
That is precisely the form you want with [math]a= 0.95\cdot 5.67\cdot 10^{-8}(h/k)[/math], b=0, c= 0, d= 1, and [math]e= -(h/k)(1- \alpha)F_{SW}- (h/k)F_{other}- T_{bot}[/math].
[math]-(h/k)(1- \alpha)F_{SW}- (h/k)F_{other}+ 0.95\cdot 5.67\cdot 10^{-8}(h/k) T_{top}^4= -T_{top}+ T_{bot}[/math].
That can be written
[math]0.95\cdot 5.67\cdot 10^{-8}(h/k) T_{top}^4+ T_{top}+ -(h/k)(1- \alpha)F_{SW}- (h/k)F_{other}- T_{bot}= 0[/math].
That is precisely the form you want with [math]a= 0.95\cdot 5.67\cdot 10^{-8}(h/k)[/math], b=0, c= 0, d= 1, and [math]e= -(h/k)(1- \alpha)F_{SW}- (h/k)F_{other}- T_{bot}[/math].