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Let $p(x) = ax^3 + bx^2 + cx + d$ where $a,b,c,d \in\mathbb{R}$. Show that if the curve $y^2 = p(x)$ has a double point, then it must be of the form $(r,0)$ where $r$ is a double root of $p(x)$.

$$p(r) = p'(r) = 0 \implies (x-r)^2\mid p(x)$$

$$p(x) = y^2 \implies p'(x) = 2y \text{ or } y'$$

I am stuck and don't know where or how to go from here.

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    If you are taking derivatives on both sides of $p(x)=y^2$, then the answer *should* be $p'(x) = 2yy'$, not "$p'(x)=2y$ or $p'(x)=y'$".2012-07-18

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