If the cubic polynomial with rational coefficients $x^3+ax^2+bx+c$ has a double root, the root is rational. This question comes from a problem in Rational Points on Elliptic Curves by Silverman and Tate. The problem asks to show that 

For a cubic curve $C: y^2=x^3+ax^2+bx+c$ with $a,b,c\in \mathbb{Q}$, if $S=(x_0, y_0)$ is a singular point, then $(x_0,y_0)$ is rational.

It turns out that a singular point $(x_0, y_0)$ must satisfy
$$y_0=0, -3x_0^2-2ax_0-b=0.$$
In other words, a point $(x_0,y_0)$ is singular on this cubic curve if and only if $y_0=0$ and $x_0$ is a multiple root of $x^3+ax^2+bx+c$.
If it is a triple root, it is easy to see $x_0$ has to be rational. It is in fact $-\frac{a}{3}$.
If it is a double root, we can write the above cubic polynomial as $(x-x_0)^2(x-x_1)$ where $x_1$ is another root. We have
$$2x_0+x_1=-a, x_0^2x_1=-c, x_0^2+2x_0x_1=b.$$
If we suppose $x_0$ is irrational, then $x_0^2, x_1$ have to be irrational too. But this does not help. 
How should I continue? Thank you for any help!

HINT.-Put the equation of the cubic as $$f(x)=y^2-(x-\alpha)(x-\beta)(x-\gamma)=0$$ where $\alpha,\beta,\gamma$ are supposed reals in the affin plan.
We must have $\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}=0$. 
Since $\frac{\partial f}{\partial y}=2y=2\sqrt{(x-\alpha)(x-\beta)(x-\gamma)}$ then 
$$(x-\alpha)(x-\beta)(x-\gamma)=0$$
On the other hand it is known that the cubic is non-singular if and only if $\alpha,\beta$ and $\gamma$ are distinct so making $\alpha\le\beta\le\gamma$ we have as  possibilities
$$\begin{cases}(x-\alpha)(x-\beta)^2: (\beta,0)\text{ nodal point}\\(x-\alpha)^2(x-\beta):(\alpha,0)\text{ isolated point}\\(x-\alpha)^3:(\alpha,0) \text{ cusp point}\end{cases}$$
That these points are rational is obtained from the system
$$\begin{cases}\alpha+\beta+\gamma=-a\\\alpha\beta+\alpha\gamma+\beta\gamma=b\\\alpha\beta\gamma=-c\end{cases}$$
