# Prove that $ax^3+bx^2+c = 0$ have less then three real roots on $[0,1]$

I need to know whether what I wrote below makes sense. Plus if so what can be improved.

Suppose the opposite there exist $$g(x) =ax^3+bx^2+c$$ that has 3 real roots in $[0,1]$ and it is not zero polynomial. If $a = 0$ then $deg(g)\leq2$ so it cannot have three roots.
So $a \neq 0$.

$$g(x) = ax^3 + bx^2 + c$$ There exist $\alpha_1$, $\alpha_2$, $\alpha_3$ $\in [0,1]$ such that $g(\alpha_1)=g(\alpha_2)=g(\alpha_3)=0$. Polynomial is continuous and differentiable so by Rolle theorem there exist $\beta_1 \in (\alpha_1, \alpha_2)$, $\beta_2 \in (\alpha_2, \alpha_3)$ $$f'(\beta_1) = 0, f'(\beta_2) = 0$$ $$f'(x) = 3ax^2+2bx$$ $$f'(\beta_i) = 0 = 3a\beta_i^2 + 2b\beta_i$$ $\beta_i = 0$ is a root. For $\beta_i \neq 0$ there is: $$0 = 3a\beta_i^2 + 2b\beta_i$$ $$-2b = 3a\beta_i + 2b$$ $$\frac{-2b}{3a} = \beta_i$$ $\beta_1 = 0$, $\beta_2 = \frac{-2b}{3a}$. but $\beta_1 \in (\alpha_1, \alpha_2) \subset (0,1)$

• Sorry but i have an objection . You make a logical mistake i think ... if $\beta_i$ is such that it's satisfying the Rolle theorem then it has the form $3\alpha (\beta_i)^2+ 2b\beta_i =0$ but the opposite doesn't apply , namely $3\alpha (\beta_i)^2+ 2b\beta_i =0 \Rightarrow \beta_i$ comes from Rolle theorem. – chaviaras michalis Aug 23 '16 at 8:46

A different approach can be as follows. We will argue by contradiction. Assume that there are three different roots in $[0,1],$ say $\alpha, \beta,\gamma.$ Then

$$ax^3+bx^2+c=a(x-\alpha)(x-\beta)(x-\gamma)=a[x^3-(\alpha+\beta+\gamma)x^2+(\alpha\beta+\alpha\gamma+\beta\gamma)x+\alpha\beta\gamma].$$

So, it must be $$\alpha\beta+\alpha\gamma+\beta\gamma=0.$$ Since $\alpha, \beta,\gamma\ge 0$ we get that at least two of the three roots must be zero. That is, the polynomyal has at most two different roots, what gives us the desired contradiction.

From what I see, your proof is on the correct path if the tasks specifically says three different roots (i.e., that none of the roots is double).

However, you can shorten it since you don't really need to find $\beta_2$, it's enough to show that $\beta_1=0$, and that this contradicts the fact that $\beta_1\in(0,1)$.

Also, the task would become a little easier if you assume (which you can without loss of generality) that $a=1$. If it is not, simply look at the polynimial $\frac{g(x)}{a}$

Should it be $(0,1]$ instead of $[0,1]$? Also it seems that $(0,1]$ can be replaced by $(0,+\infty)$. If $x_1$, $x_2$, and $x_3$ denote the roots, then, by Vieta's formula, $x_1x_2+x_1x_3+x_2x_3=0$. Hence, if all roots are non-negative, then at least two of them are zero...