Hello, topsquark!
Spoiler:
Solve for x:
without using the rational root theorem.
Hint: There is one integer solution.
There is a way to factor this but if you actually want to slog through it using Cardano's Method I'll give you a hero cookie.
-Dan
Did you not understand topsquark's post? That is NOT Cardano's method so: NO cookie for you!
However, topsquark, Cardano's method solve the "reduced cubic" of the form . Of course, any cubic can be put in that form but did you mean ?
The problem is correct as written. Of course, as you say, you can change the form of the equation to put it into x^3 + mx + q by a translation of coordinates and in that form it may be "attacked" ala Cardano. I was hoping that no one would use it, but instead apply counter-terms to the problem.
Am I missing something? Soroban's solution is correct and to the best of my knowledge doesn't use Cardano's method?
-Dan
yeah we can check check the function for monotonicity
f'(x) should not be equal to 0 in (a,b) and f''(x)>=0 or <=0 for all x in (a,b) (for concavity)
at endpoints a,b mod(f(a)/f'(a))<b-a and mod(f(b)/f'(b))<b-a
i think all these conditions are holding for above