Two points on curve that have common tangent line Find the two points on the curve $y=x^4-2x^2-x$ that have a common tangent line.
My solution: Suppose that these two point are $(p,f(p))$ and $(q,f(q))$ providing that $p \neq q$. Since they have a common tangent line then: $y'(p)=y'(q),$ i.e. $4p^3-4p-1=4q^3-4q-1$ and after cancellation we get: $p^2+pq+q^2=1$.
Tangent lines to curve at points $(p,f(p))$ and $(q,f(q))$ are $y=y(p)+y'(p)(x-p)$ and $y=y(q)+y'(q)(x-q)$, respectively. I have tried to put $x=q$ in the first and $x=p$ in the second equations but my efforts were unsuccesfull.
Can anyone explain me how to tackle that problem?

Complete the square :
your curve is $y = x^4 - 2x^2 - x = (x^2 - 1)^2 + (-x-1)$. 
So the curve stays above the line $y = -x-1$, and is tangent to it when it touches it, that is when $x^2-1 = 0$, so that's when $x=1$ and $x=-1$ :
The line $y=-x-1$ is tangent to the curve at those two points $(1,-2)$ and $(-1,0)$.
