Find the value of a + b + c + d Let $a$ and $b$ be the roots of the equation: $x^2 - 10cx - 11d = 0$ where $c$ and $d$ be the roots of $x^2 - 10ax - 11b = 0$. Find the value of $a+b+c+d$, assuming that they all are distinct.
I first tried making an equation with roots $(a+b)$ and $(c+d)$ to get the sum of the roots, however I wasn't able to solve this question using that method as the answer which I got was in terms of the variables itself.
I also tried placing $a$ into the first equation and $c$ into the second to cancel out a common term ($-10ac$), but after cancelling, I got: $(a^2 - c^2 - 11d + 11b = 0)$. Now I don't know how to move ahead.
 A: My Solution ::
Given $a,b$ are the roots of $x^2-10cx-11d=0.$
So $$ a+b = 10c \tag{1}$$
and
$$ab = -11d \tag{2} $$
and $c,d$ are the roots of $x^2-10ax-11b=0.$
So $$c+d=10a\tag{3}$$
and
$$cd=-11b \tag{4}$$
So $$a+b+c+d = 10(a+c). \tag{5}$$
Now $$\frac{a+b}{c+d} = \frac{10c}{10a}=\frac{c}{a}\Rightarrow a^2+ab=c^2+cd\Rightarrow a^2-11d=c^2-11b$$
So $$(a^2-c^2)=-11(b-d)\Rightarrow (a+c)\cdot(a-c)=-11(b-d) \tag{6}$$
Now $(1)-(3)$, we get $$a+b-c-d=10c-10a\Rightarrow (b-d) = 11(c-a)=-11(a-c)$$
Now putting $(b-d) = -11(a-c)$ in eqn. $(6)$, we get $$(a+c)\cdot(a-c)=121(a-c)$$
So $(a-c)\cdot(a+c-121) = 0$.  Now $a\neq c$, because $a,b,c,d$ are distinct real numbers.
So $a+c=121$.  Put into eqn. $(5)$.
We get
$$
a+b+c+d = 10(a+c) = 10\cdot 121 = 1210\Rightarrow (a+b+c+d) = 1210.
$$
A: We have
$a+b=10c$ 
and
$c+d=10a$, hence
$$b=10c-a$$
and
$$d=10a-c$$
We also have $ab=-11d$ and $cd=-11b$, hence
$$10ac-a^2=-11(10a-c)$$
and
$$10ac-c^2=-11(10c-a)$$
Subtracting these gives
$$a^2-c^2=-11(10c-a)+11(10a-c)=121(a-c)$$
Assuming $a\not=c$, we get
$$a+c=121$$
Going back to the formulas for $b$ and $d$, we have
$$b+d=(10c-a)+(10a-c)=9(a+c)$$
Thus
$$a+b+c+d=(a+c)+(b+d)=(a+c)+9(a+c)=10(a+c)=1210$$
