2016/028) Let $a,b$ be roots of equation

Let $a,b$ be roots of equation $x^2-10cx+11d=0$ and $c,d$ be roots of equation
$x^2-10ax-11b=0$ then find the value of $a+b+c+d$ when a,b,c,d are all distinct.

we have $a,b$ are roots of $x^2-10cx + 11d = 0$
Hence $a+b = 10c\cdots(1)$
$ab= 11d\cdots(2)$
similarly
$c+d= 10a\cdots(3)$
$cd = 11b\cdots(4)$
from (1) and  (3)
$a+b+c+d = 10(c+a)$ or $b+d = 9(a+c)\cdots(5)$
from (2) and (4)
$abcd = 121bd$ or $ac = 121\cdots(6)$
now because a and b satisfy $x^2-10cx + 11d = 0$
we have
$a^2-10ca + 11d = 0$
similarly
$c^2 10ca + 11b = 0$
add to get $(a+c)^2 -2ac - 20ac +11(b+d) = 0$
or $(a+c)^2 - 22 * 121 + 99(a+c) = 0$
So $a + c = - 22\, or\,  121$
$a + c = -22$ means $a = c = -11$  wich is inadmissible
so $a + c = 121$
so $a + b+ c+ d = 10(a+ c) = 1210$