Strange Vieta's formula

Make $x=y+1$ to get a new equation with no cube term: $y^4-22y^2-48y-23=0$ . Then, factor it into the form $(y^2+py+q)(y^2-py+r)=0$ . Expanding and comparing coefficients we get the system $q+r=p^2-22$ , $q-r=48/p$ and $qr=-23$ . Using the identity $(q-r)^2=(q+r)^2-4qr$ we get $p^6-44p^4+576p^2-2304=0$ , which factors as $(p^2-24)(p^2-12)(p^2-8)=0$ .

Now, back to our problem we see that, by Vieta's formula, $q=(a-1)(b-1)$ and $r=(c-1)(d-1)$ . Adding that we get $q+r=ab-a-b+1+cd-c-d+1$ or $p^2-22=ab+cd-(a+b+c+d)+2$ or $ab+cd=p^2-20$ . Since we have three possible values for $p^2$ we also have three possible values for $ab+cd$ , which are $4,-8,-12$ . Adding their absolute values we get $\boxed{24}$ .

The above equation can be transformed into

$(x^{2} -2x+2)^{2} = 24x^{2}$ .

Now from here we get two quadratic equations with roots $(x,y), (p,q)$ , respectively.

$x^{2} -2x( -\sqrt{6}+1) + 2$

And

$x^{2} -2x( \sqrt{6}+1) + 2$ .

Now sum of all absolute values of $ab+cd$

= $\boxed{24}$