Real points of order 3 on an elliptic curve. This comes from Silverman's Rational Points on Elliptic Curves:
Consider the elliptic curve (non singular) $y^2=x^3+ax^2+bx+c=f(x)$ after some computations we can see that points of order 3 in this curve satisfy $\psi_3(x))=0$ $\psi_3(x)=2f(x)f''(x)-f'(x)^2=3x^4+4ax^3+6bx^2+12cx+(4ac-b^2)$
I want two show that there are only two real roots of $\psi_3$ if a,b,c $\in \mathbb{R}$:
First $\psi_3$ goes to infinity if $x \rightarrow \pm \infty$ and $f''(\frac{-a}{3})=0 \implies \psi_3(\frac{-a}{3})<0$ by continuity we get $\alpha_1$, $\alpha_2$ $\in \mathbb{R}$ roots of $\psi_3$. What do I do to show that these are the only real roots?

Let $E: y^2=f(x)$ and $\psi_3(x) = 2f(x)f''(x)-(f'(x))^2$, where $f(x)$ is a monic cubic polynomial with three distinct roots (because $E$ is non-singular!), as above. This can be shown using the following hint:


*

*A quartic polynomial $p(x)=a_4x^4+\cdots+a_0$, with $a_4>0$, has exactly two real roots if  $p(x)$ takes negative values at all the zeros of the derivative $p'(x)$.


Thus, you just need to show that $\psi_3(x)$ is negative at the zeros of $\psi_3'(x)$, using the fact that $\psi_3'(x)$ has a simple expression in terms of $f(x)$. You also need to use the fact that $f(\alpha)=f'(\alpha)=0$ is not possible under the above hypotheses.
