Second point of intersection - parameter removes second solution? $$\mathbf r_1 (t) = (t^2 - t, t^2 + t) \\ \mathbf r_2(u) = (u+u^2, u-u^2)$$
I'm trying to find two of the intersection points, but I'm lost as to how to approach the question. Is it possible to remove the parameter?
$$\mathbf r_1 - \mathbf r_2 = (t^2-t-u-u^2)\mathbf{\hat i} + (t^2+t-u+u^2)\mathbf{\hat j} = \mathbf0 $$So,
$$ t^2 -t - u - u^2 = 0
\\t^2+t-u+u^2 = 0
$$
Solving, $$ t=0,u=0 \\ t=-1, u = 1$$
When I sub back into the equations, I don't get the same points for $t=-1, u =1$, so I'm doing something wrong...The answers say intersection points are $(0,0)$ and $(2,0)$. Where does $(2,0)$ come from?
 A: You did have the right idea: your system of equations produces
$$  \begin{array}{cc}t^2 -t - u - u^2 = 0\\t^2+t-u+u^2 = 0\end{array} \ \Rightarrow \ 2t^2  - \ 2u \ = \ 0 \ \Rightarrow \ t^2 \ = \ u \ , $$
which you appear to have replaced into one or the other coordinate equations,
$$ t^2 - t  \ = \ u + u^2 \ \Rightarrow \ t^2 - t  \ = \ t^2 + t^4  \ \Rightarrow \ t \cdot ( t^3 + 1 ) = 0 \ \ \ \text{or} $$
$$ t^2 + t  \ = \ u - u^2 \ \Rightarrow \ t^2 + t  \ = \ t^2 - t^4  \ \Rightarrow \ t \cdot ( t^3 + 1 ) = 0 \ , $$
which has only two real solutions, $ \ t = 0 \ \ \text{and} \ \ t = -1 \ . $  These correspond to solutions for the other parameter, $ \ u = 0 \ \ \text{and} \ \ u = +1 \ . $   
You found this successfully.  Inserting these parametric values into your vector equations produce
$$\vec{r_1}(0) \ = \ (0^2 - 0 \ , \ 0^2 + 0 ) \ = \  (0 , 0 ) \ = \ \vec{r_2}(0)  $$
and
$$\vec{r_1}(-1) \ = \ ([-1]^2 - [-1] \ , \ [-1]^2 + [-1] )  \ = \ (2, 0)$$
$$ ([+1]^2 + [+1] \ , \ [+1]^2 - [+1] ) \ = \  \vec{r_2}(+1) \ , $$
the latter being an intersection point corresponding to distinct values of the parameters $ \ t \ $ and $ \ u \ $ , as Tony Piccolo mentions in his comment.
Here is a graph of the situation:

A: I begin supposing that the curves $r_1(t), r_2(t)$ are both parametrized by the same parameter $t$ in the interval $[a,b]$, with $0\in[a,b]$. Then $(0,0)$ is the unique point of intersection of the curves $r_1(t)=(t^2-t,t^2+t)$ and $r_2(t)=(t^2+t,-t^2+t)$ as $\langle r_1(t), r_2(t)\rangle=0$, for all $t\in[a,b]$. Equivalently, the eqs. $t^2-t=t^2+t$ and $-t^2+t=t^2+t$ admit $t=0$ as only solution. 
If $t\in[a,b]$ but $u\in[c,d]$ with $[a,b]\neq[c,d]$, then we can introduce the variable $\bar{u}\in[a,b]$ with the transformation
$\bar{u}=\frac{ad-bc}{d-c}+\frac{b-a}{d-c}u:=\alpha+\beta u$, 
or equivalently $u=\frac{\bar{u}-\alpha}{\beta}$, and re-parametrize $r_2$ as follows:
$$[a,b]\ni\bar{u}\mapsto r_2(\bar{u})=((\frac{\bar{u}-\alpha}{\beta})^2+(\frac{\bar{u}-\alpha}{\beta}),(\frac{\bar{u}-\alpha}{\beta})-(\frac{\bar{u}-\alpha}{\beta})^2)$$
Now you can call $\bar{u}=t$ in $r_2(\bar{u})$ as it is a dummy variable (the important thing it is that $\bar{u}\in[a,b]$ as $t$) and try to solve  $r_1(t)=\tilde{r}_2(t)$ with
$\tilde{r}_2(t)=((\frac{t-\alpha}{\beta})^2+(\frac{t-\alpha}{\beta}),(\frac{t-\alpha}{\beta})-(\frac{t-\alpha}{\beta})^2)$,
though. The solution(s) depends on $a,b,c,d$, in general.
