<s>[INST] Show that the curve $y^2 = x^3 + 2x^2$ has a double point, and find all rational points Show that the curve $y^2 = x^3 + 2x^2$ has a double point. Find all rational points on this curve.
By implicit differentiation of $x$, $-3x^2 - 4x$ vanishes iff $x = -4/3$ and $0$.
By implicit differentiation of $y$, $2y$ vanishes iff $y = 0$.
Taking the second derivative, I got $-6x-4$ and then using the point on the curve $(0,0)$ I got $-4$. Is this my double point?
Thanks for any help!
 [/INST] Since you are a visual person, you should try drawing it. It seems fairly obvious from the real picture that there is a double point at zero, and then you just use whatever methods your book usually employs to prove that it actually is a double point. 
On the other hand, finding all rational solutions means finding all rational points such that 
$$y = \pm\sqrt{x^3+2x^2} = \pm~ x\sqrt{x+2}.$$
I will omit the sign for the rest of this post. Note that $x+2$ must have a rational root. In other words, $x+2=\frac{p^2}{q^2}$ for integers $p$ and $q$, so $x=\frac{p^2-2q^2}{q^2}$. This yields
$$y = \frac{p^2-2q^2}{q^2} \cdot \frac{p}{q}$$
and multiplying by $q^3$, we obtain that
$$q^3y = p^3 - 2pq^2$$
over the integers. Hence, any rational solution of your equation is parametrized by $p,q\in\mathbb{Z}$ and then,
$$\begin{align*}
y &= \frac{p^3-2pq^2}{q^3} \\
x &= \frac{p^2-2q^2}{q^2}.
\end{align*}$$
</s>