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lang-mma
(2*x1^2 + 3*x1*x2 + 2*x2^2)/(4 x1^2*x2 + 4*x2^2*x1) /. Thread[{x1, x2} -> SolveValues[2 x^2 + 5 x - 7 == 0, x]]. What would you want to go any other way? $\endgroup$2 x1^2 + 5 x1 - 7 == 0 && 2 x2^2 + 5 x2 - 7 == 0 && x1 < x2are enough for the simplification, but it is not so. Also think about the analog of the question for cubic, say to simplify a symmetric expression formed byx1,x2,x3which are rhe roots of a cubic. Don't hesitate to ask forv further explanation in need $\endgroup$(2*x1^2 + 3*x1*x2 + 2*x2^2 + 3 x2*x3 + 2 x3^2 + 3 x1*x3)/(4 x1^2*x2 + 4 x1^2*x3 + 4*x2^2*x1 + 4 x2^2*x3 + 4 x3^2*x1 + 4 x3^2*x2) /. Thread[{x1, x2, x3} -> SolveValues[2 x^3 + 5 x - 7 == 0, x]]outputs(2 + 3/2 (-1 - I Sqrt[13]) + 1/2 (-1 - I Sqrt[13])^2 + 3/2 (-1 + I Sqrt[13]) + 3/4 (-1 - I Sqrt[13]) (-1 + I Sqrt[13]) + 1/2 (-1 + I Sqrt[13])^2)/(2 (-1 - I Sqrt[13]) + (-1 - I Sqrt[13])^2 + 2 (-1 + I Sqrt[13]) + 1/2 (-1 - I Sqrt[13])^2 (-1 + I Sqrt[13]) + (-1 + I Sqrt[13])^2 + 1/2 (-1 - I Sqrt[13]) (-1 + I Sqrt[13])^2). $\endgroup$16/35$\endgroup$