1. ## Finding number of roots of third grade eqautions

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

2. Originally Posted by hejmh
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
Dear hejmh,

$2x^3-3x^2+1+a=0$

This is a polynomial equation of degree three. Therefore by the fundamental theorem of algebra there are three roots. It does'nt matter whatever value you choose for a.

3. Originally Posted by Sudharaka
This is a polynomial equation of degree three. Therefore by the fundamental theorem of algebra there are three roots. It does'nt matter whatever value you choose for a.
I suppose he means "real roots". We can draw the graph of:

$f(x)=2x^3-3x^2+1$

and observe the number of intersection points with $y=-a$.

Fernando Revilla

4. I changed the $a$ to $\lambda$ for convenience. (Also, this most probably isn't the way you're supposed to do).
Originally Posted by hejmh
How many roots does this equation have and how does the number of roots change after changing $\lambda$? $2x^3-3x^2+1+\lambda=0$
The discriminant of the cubic $ax^3+bx^2+cx+d$ is given by:

$\Delta=b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd$

In our case, all the terms that contain $c$ vanish, so we're left with:

$\Delta = -4b^3d-27a^2d^2 = -4(-3)^3(1+\lambda)-27(2)^2(\lambda+1)^2$

$= 108 (\lambda+1)-108 (\lambda+1)^2 = -108 \lambda (\lambda+1).$

For $2x^3-3x^2+1+\lambda=0$ to have (all) real solutions, we must have:

$-108 \lambda (\lambda+1) \ge 0 \Rightarrow \lambda (\lambda+1) \le 0 \Rightarrow \boxed{\boxed{-1 \le \lambda \le 0}}.$

5. Complex roots always occur as conjugates. So that means that there can either be 3 real roots, or 1 real and 2 complex roots.

6. Originally Posted by Prove It
Complex roots always occur as conjugates. So that means that there can either be 3 real roots, or 1 real and 2 complex roots.
Of course, but I think the question is for what values of $a$ does the given polynomial
have have three real solutions or one real and two complex solutions!

7. Sometimes a picture helps.

(1 + a) is a constant value that shifts the graph of y = 2x^3-3x^2 up and down. What happens when a is non-zero and non-negative?

8. We can consider the function f defined by : f(x)=2x^3-3x^2+1+a on R show that it's increasing on ]-00,0]U[1,+00[ and decreasing on [0,1] and use the fact that it's a bijection , or we can use the Intermediate value theorem and discuss the values of a .

9. 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 $2x^3- 3x^2+ a+ 1= 0$, 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. $2(-x)^3- 3(-x)^2+ a+ 1= -2x^3- 3x^2+ a+ 1= 0$. 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 $2x^3- 3x^2= x^2(2x- 3)= 0$ which has x= 0 as a double root and x= 3/2 as a positve root.