Editing SR/Ptot QuarticSolution (section)

===Quartic Equation Solution===

Following the [http://en.wikipedia.org/wiki/Quartic_function#Summary_of_Ferrari.27s_method Summary of Ferrari's method] that is presented in Wikipedia's discussion of the Quartic Function to identify the roots of an arbitrary quartic equation, we can set the coefficients of the cubic and quadratic terms both to zero and deduce that the only root that will give physically relevant temperatures &#8212; for example, real and non-negative &#8212; is,
<div align="center">
<math>
2z = \biggl[\frac{2a_1}{a_4 W}-W^2\biggr]^{1/2} - W ,
</math>
</div>
where,
<div align="center">
<math>
\frac{1}{2}W^2  \equiv R^{-1/3}\biggl[R^{2/3} - \frac{a_0}{3a_4}  \biggr] ,
</math><br/>
<math>
R \equiv \biggl( \frac{a_1}{4a_4} \biggr)^2 \biggl[ 1 + (1 + \lambda^3)^{1/2} \biggr] ,
</math><br />
<math>
\lambda^3 \equiv \frac{2^8 a_0^3 a_4}{3^3 a_1^4} .
</math>
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Defining,
<div align="center">
<math>
\phi \equiv \frac{2a_1}{a_4 W^3} ~~~~~\Rightarrow ~~~~~ W = 2\biggl(\frac{a_1}{4a_4 \phi} \biggr)^{1/3} ,
</math>
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and realizing that, from one of the above expressions, 
<div align="center">
<math>
\frac{1}{2}W^2  = \biggl[ \frac{a_1^4}{2^8 a_4^4 R}\biggr]^{1/3}\biggl[\biggl( \frac{2^4 a_4^2 R}{a_1^2}\biggr)^{2/3} - \biggl( \frac{2^8 a_0^3 a_4}{3^3 a_1^4} \biggr)^{1/3}  \biggr]
= \biggl[ \frac{a_1}{2^2 a_4}\biggr]^{2/3} \biggl[ 1 + (1 + \lambda^3)^{1/2} \biggr]^{-1/3}\biggl\{ \biggl[ 1 + (1 + \lambda^3)^{1/2} \biggr]^{2/3} - \lambda \biggr\} ,
</math><br/>
</div>
we can rewrite the desired root of our quartic equation in the form,
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<math>
z = \biggl(\frac{a_1}{4a_4}\biggr)^{1/3} \phi^{-1/3}\biggl[ (\phi - 1)^{1/2} - 1 \biggr] ,
</math>
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with,
<div align="center">
<math>
\phi = 2^{3/2} \biggl[ 1 + (1 + \lambda^3)^{1/2} \biggr]^{1/2}
\biggl\{ \biggl[ 1 + (1 + \lambda^3)^{1/2} \biggr]^{2/3} - \lambda \biggr\}^{-3/2} .
</math>
</div>