# Cubic equation roots

1. Dec 17, 2006

### littleHilbert

1. The problem statement, all variables and given/known data

Show that for negative c (a,b,c - real) equation $x^3+ax^2+bx+c=0$ has at least one positive root.

2. The attempt at a solution

Considering the equivalent form of the equation above for large |x|:
$x^3(1+\frac{a}{x}+\frac{b}{x^2}+\frac{c}{x^3})=0$ we can conclude that there exists at least one real root, because the function changes signes if x does the same.

Let r,s,t be some roots of this equation. At least one of them is real. The cubic polynomial on LHS can be written in the form:
$(x-r)(x-s)(x-t)$. Thus the equation admits the following form: $x^3-(r+s+t)x^2+(rs+st+rt)x-rst=0$. Comparing yields: -(r+s+t)=a, rs+st+rt=b and -rst=c. Now if for example r is real and so are s and t, then in order for c to be negative the product rst must be positive. This in turn is true only if at least one the factors is positive.

Is it correct?

Now the task says nothing about whether all the roots are assumed to be real or not. In case only one root is real, for example r, is it correct to say that the product st is real?

Is there another way to solve the problem?

2. Dec 17, 2006

### cristo

Staff Emeritus
If all are not real, then consider -rst=c. Supposing r is real, as above, then s and t must necessarily be complex conjugates of each other (otherwise the product rst would be complex, which is not allowed since c is real) , and so st>0, which gives r>0

3. Dec 17, 2006

### Hurkyl

Staff Emeritus
There's a good approach to this problem using calculus.

4. Dec 17, 2006

### AlephZero

You were getting close when you started thinking about changes of sign of the function. A polynomial is a continuous function, so if you can show it changes sign in some interval [0,X] for some (large) value of X, it has a root in that interval.

5. Dec 21, 2006

### HallsofIvy

Staff Emeritus
Yes, he understood that. That's why he asked "is this another way to solve the problem".

6. Dec 31, 2006

### Gib Z

Mean Value Theorem.