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Bisection Method to solve a polynomial?

Posted 10 years ago
POSTED BY: Jo Ko

In Mathematica it would be something like this.

FindRoot[x^2 - 7 == 0, {x, 0, 10}, Method -> "Brent"]

(* Out[22]= {x -> 2.64575} *)

I don't know if Brent's method is exactly bisection though. Maybe use the secant method?

FindRoot[x^2 - 7 == 0, {x, 0, 10}, Method -> "Secant"]

(* Out[25]= {x -> 2.64575} *)
POSTED BY: Daniel Lichtblau
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