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:
$\displaystyle f(x)=2x^3-3x^2+1$
and observe the number of intersection points with $\displaystyle y=-a$.
Fernando Revilla
I changed the $\displaystyle a$ to $\displaystyle \lambda$ for convenience. (Also, this most probably isn't the way you're supposed to do).
The discriminant of the cubic $\displaystyle ax^3+bx^2+cx+d $ is given by:
$\displaystyle \Delta=b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd$
In our case, all the terms that contain $\displaystyle c$ vanish, so we're left with:
$\displaystyle \Delta = -4b^3d-27a^2d^2 = -4(-3)^3(1+\lambda)-27(2)^2(\lambda+1)^2 $
$\displaystyle = 108 (\lambda+1)-108 (\lambda+1)^2 = -108 \lambda (\lambda+1). $
For $\displaystyle 2x^3-3x^2+1+\lambda=0$ to have (all) real solutions, we must have:
$\displaystyle -108 \lambda (\lambda+1) \ge 0 \Rightarrow \lambda (\lambda+1) \le 0 \Rightarrow \boxed{\boxed{-1 \le \lambda \le 0}}.$
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 $\displaystyle 2x^3- 3x^2+ a+ 1= 0$, 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. $\displaystyle 2(-x)^3- 3(-x)^2+ a+ 1= -2x^3- 3x^2+ a+ 1= 0$. 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 $\displaystyle 2x^3- 3x^2= x^2(2x- 3)= 0$ which has x= 0 as a double root and x= 3/2 as a positve root.