Prove that for a cubic of the form then there exists at least one set of real solutions such that:
Strictly speaking, what you want to prove here is NOT true! If and , the first two conditions are incompatible. I am going to assume that .
Since that third condition follows from the first so we really have only two conditions. That means we are free to choose one of the parameters to be whatever we want. Choosing makes the first condition . With and , the second condition becomes which gives as long as .
If we have and , then we really only have one condition. Choose and . a now can be any number at all.
hey this comes under theory of equations,right??? i expanded (x-b)^3 and got a cubic. if the roots be y1,y2,y3 then i got the following:
y1+y2+y3=-3b
y1y2+y2y3+y3y1=-3b^2
y1y2y3=d/a-b^3
as evident there should be 3 roots but i simply couldnot find y1,y2,y3.there seems to be some kind of a proof not by usual methods.