This simple Solve gives the roots of a quadratic:
Solve[a x^2 + b x + c == 0, x]
However, if I factor the polynomial in terms of its "unknown" roots x1 and x2, this code does not work
Solve[a x^2 + b x + c == (x - x1) (x - x2), {x1, x2}]
to solve for x1 and x2 (which in this case, we know, they are the roots found by the first code. Why not? How can I code a solution in this spirit in mathematica?
I need to find a way to employ a solution which is in the spirit of the second code, however, for the following reason.
In my real problem, I seek the roots of a degree 5 polynomial. Indeed, this general problem cannot be solved (see: Galois). However, in my case, several of the roots are given by known functions of the other roots. I plan to insert this information into the factored form on the RHS of code in the form of my second code above.
Answer
Reduce[ForAll[x, a x^2 + b x + c == a (x - x1) ( x - x2)], {x1, x2},Backsubstitution -> True]
gives you the solution you want.
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