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Given that the solution set of inequation

$$ax ^ 2-bx-1>0$$ is $-1/2<x<-1/3$, I want to calculate the solution set of inequation

$$x ^ 2-bx-a\geq 0.$$

I have used the following code, but is there a faster and better way to solve this problem?

Clear["Global`*"]
sol = First@Solve[{-1/2 - 1/3 == b/a, 1/6 == -1/a}, {a, b}]
(* {a -> -6, b -> 5} *)

Reduce[x^2 - b x - a >= 0, x] /. sol
(* x ∈ Reals && (x <= 2 || x >= 3) *)
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  • $\begingroup$ From where do you obtain {-1/2 - 1/3 == b/a, 1/6 == -1/a}? $\endgroup$
    – user64494
    Commented Aug 15, 2023 at 6:51
  • $\begingroup$ Obviously, these two numbers are the two roots of the equationax ^ 2-bx-1=0 $\endgroup$
    – csn899
    Commented Aug 15, 2023 at 6:59

1 Answer 1

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First, we find the set of a and b for which the solution of a*x ^ 2-b*x-1>0 is x > -1/2 && x < -1/3.

cond = Resolve[ForAll[x, x > -1/2 && x < -1/3, a*x^2 - b*x - 1 > 0] && 
ForAll[x, x >= -1/3, a*x^2 - b*x - 1 <= 0] && 
ForAll[x, x <= -1/2, a*x^2 - b*x - 1 <= 0], Reals]

A long output, its LeafCount equals 5460

Now

Reduce[x^2 - b*x - a >= 0 && cond, x, Reals]

b == 5 && a == -6 && (x <= 2 || x >= 3)

finishes the work.

Edit. ForAll instead of Exists in cond = Resolve[ForAll[x, x > -1/2 && x < -1/3, a*x^2 - b*x - 1 > 0] && Exists[x, x >= -1/3, a*x^2 - b*x - 1 <= 0] && Exists[x, x <= -1/2, a*x^2 - b*x - 1 <= 0], Reals]

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