@StevenLandsburg Thanks! Formula 7.4.13 in the chapter you cite gives $x=(c+d_1-d_2)/2$ when $F(x)=(x-d_1)(c-x-d_2)$, confirming that Nash's result subsumes the particular case I discussed in ¶5. If someone wishes to write a short summary including of this theorem (for completeness of MathOverflow) as an answer, I will mark it as accepted.
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