Rooting for the roots

Applying sum of roots and product of roots,

$a+b=10c; c+d=10a$ - (i)

$ab=-11d; cd=-11b$ - (ii)

Adding both the equations in (i), we get $a+b+c+d=10(a+c) \rightarrow b+d=9(a+c)$

Equating the value of $\frac {b}{d}$ from (ii), we get $ac=121$ . Now,

$a^2-10ac-11d=0$

$c^2-10ac-11b=0$

Adding both the equations above, we get,

$((a+c)^2-2ac)-20ac-11(b+d)=0$

Putting $a+c=t, b+d=9t, ac=121$ , we get a quadratic equation in $t$ .

$(t-121)(t+22)=0$ , $t$ = 121, -22.

But, if we make $t=-22$ , that would mean $a+c=-22;ac=121 \rightarrow a=c=-11$ , but that is not allowed.

Hence, $a+c=121$

$a+b+c+d=10(a+c)=1210$