$2$ points on a curve have a common tangent 
Let $2$ points $(x_1,y_1)$ and $(x_2,y_2)$ on the curve $y=x^4-2x^2-x$ have a common tangent line. Find the value of $|x_1|+|x_2|+|y_1|+|y_2|$.

It seems to me that I a missing a link and hence the problem appears incomplete to me. Is there something which I might be missing? Thanks.

HINT:
The equation of the tangent at $(t,t^4-2t^2-t)$ will be
$$\dfrac{y-(t^4-2t^2-t)}{x-t}=4t^3-4t-1$$
$$x(4t^3-4t-1)-y=3t^4-2t^2$$
So, the equation of the tangents at $(x_1,y_1),(x_2,y_2)$ will respectively be
$$x(4x_1^3-4x_1-1)-y=3x_1^4-2x_1^2\ \ \ \ (1)$$
$$x(4x_2^3-4x_2-1)-y=3x_2^4-2x_2^2\ \ \ \ (2)$$
These two will represent the same straight line iff
$$\dfrac{4x_1^3-4x_1-1}{4x_2^3-4x_2-1}=\dfrac{-1}{-1}=\dfrac{3x_1^4-2x_1^2}{3x_2^4-2x_2^2}$$ 
$$\implies4x_1^3-4x_1-1=4x_2^3-4x_2-1\text{ and }3x_1^4-2x_1^2=3x_2^4-2x_2^2$$ with $x_1\ne x_2$
Can you take it home from here?
