A quartic identity?

Let the roots be $a,b,x,y$ and $xy=-69$

Using Vieta's formulas, we obtain the following system of equations:

$xyab=-2001$

$xy(a+b)+ab(x+y)=-269$

$xy+ab+(a+b)(x+y)=k$

$x+y+a+b=11$

Hence, from the first equation, $ab=29$ and the system is equivallent to:

$29(x+y)-69(a+b)=-269$

$(x+y)+(a+b)=11$

$k+40=(x+y)(a+b)$

Writting $x+y=u$ and $a+b=v$ , we obtain:

$29u-69v=-269$

$u+v=11$

Solving this system we obtain: $u=5, v=6$ and $k+40=uv=30$ . Then, finally:

$\boxed{k=-10}$