$2$ 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?

The line $y=ax+b$ has $2$ tangent points to the curve $y = x^4-2x^2-x$ if and only if $( x^4-2x^2-x)- (ax+b)$ has two (real) double roots $x_1,x_2$, so this polynomial has to be a perfect square $((x-x_1)(x-x_2))^2$

Can you complete the square $(x^4-2x^2+?x+?) = (x^2+?x+?)^2$ and then find its roots $x_1,x_2$ ?