I would be grateful if anybody could help me on solving this problem:
How many roots does this equation have and how does the number of roots change after changing a?
2x^3-3x^2+1+a=0
I suppose he means "real roots". We can draw the graph of:
and observe the number of intersection points with .
Fernando Revilla
You could also use "DesCarte's rule of signs"- The number of positive real roots of a polynomial is no larger than the number of sign changes as you go from highest degree to lowest and differs from it by an even number".
In , there are either one or two sign changes depending upon whether a+ 1 is positive or negative. If a< 1, so that a+ 1 is negative, the signs are "+, -, -" so there is one sign change and exactly one positive root. If a> 1, so that a+ 1 is positive, the signs are "+, -, +" so there are two sign changes and two positive roots or none.
For negative roots, replace x with -x and do the same. . Now, if a+ 1> 0, there is one sign change so the original equation has exactly one negative root. If a+1< 0, there are no sign changes so the original equation has no negative roots.
Thus, we can say that if a> -1, there is one negative root and either 0 or 2 positive roots. If a<-1, there is one positive root and no negative roots.
Of course, if a= 1, the equation is which has x= 0 as a double root and x= 3/2 as a positve root.