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Find the point where equations $x=t^2-t$ and $y= t^3 -3t-1$ cross itself.

This's the first time I meet this kind of problem, can someone give me some idea? Thank you.

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which mean the two solutions of $t^2-t = p$ are solutions to $t^3-3t-1=q$,for some $p,q$. –  Yimin Mar 23 '13 at 4:14
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3 Answers 3

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Well, what you basically need to do is find $t_1$ and $t_2$ such that $x(t_1)=x(t_2)$ and $y(t_1)=y(t_2)$, with $t_1 \neq t_2$.

To help you a little further - if you examine one of the two equations, you'll get a polynomial in $t_1$ and $t_2$ that must be divisible by $t_1-t_2$. Divide out that factor and you'll have a lower-order polynomial - solve for either $t_1$ or $t_2$, and substitute the result into the other equation. Then you'll end up with a polynomial in just one of the two values of $t$.

One solution must be the solution to $t_1=t_2$, so factor that solution out. Now solutions will exist in pairs ($t_1=a$, $t2=b$ and $t_1=b$, $t_2=a$), so it should be fairly easy to find them from there.

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There are some $p,q$ such that

$t^2-t = p$

$t^3-3t-1=q$

has two different solutions $t_1,t_2$,

consider the third solution to the second equation as $t_3$, then $t_1+t_2+t_3 = 0$[why?]. And from the first equation $t_1+t_2 = 1$, thus $t_3 =-1$, plug into the second equation,

$(-1)^3-3*(-1)-1 = q = 1$,

then $t^3-3t-2 = (t+1)(t-2)(t+1)$, then $t_1 =2, t_2=-1$, therefore $p = 2$.

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As Glen has identified $t_2\ne t_1$

$$t_2^2-t_2=t_1^2-t_1\implies(t_2-t_1)(t_2+t_1)=t_2-t_1$$ $$\implies t_2+t_1=1\text{ as } t_2\ne t_1$$

$$t_2^3-3t_2-1=t_1^3-3t_1-1\implies t_2^3-t_1^3=3(t_2-t_1)$$ $$\implies t_1^2+t_1t_2+t_2^2=3\text{ as } t_2\ne t_1$$ $$\implies (t_1+t_2)^2-t_1t_2=3\implies t_1t_2=-2$$

So, $t_1,t_2$ are the roots of the Quadratic Equation $y^2-(1)y+(-2)=0\implies y=\frac{1\pm\sqrt{1^2-4(1)(-2)}}2=\frac{1\pm3}2=2,-1$

So, $t_1,t_2$ are $2,-1$

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