Zeroes of polynomials and their sum Let $a, b$ are zeroes of the polynomial $x^2-10cx-11d$ and $c,d$ are the zeroes of the polynomial  $x^2-10a x-11b $ where $a,b,c, d$ are distinct reals then $a+b+c+d=?$
 A: K.Dutta gave you the good track to follow.
From the first equation, we have $$a+b=10c\  \ \ \  (1)$$ $$a b=-11d\  \ \ \  (2)$$ From the second equation, we have $$c+d=10a\  \ \ \  (3)$$ $$c d=-11b\  \ \ \  (4)$$ Use the first and the third of these to eliminate $c$ and $d$. This gives $$c=\frac{a+b}{10}$$ $$d=\frac{99 a-b}{10}$$ Now $$0=ab+11d=a \left(b+\frac{1089}{10}\right)-\frac{11 b}{10}$$ from which we can extract $b$ $$b=\frac{1089 a}{11-10 a}$$ Now, replacing in $cd+11b=0$ we get $$\frac{99 a^4-10890 a^3-119790 a^2+131769 a}{(11-10 a)^2}=0$$ By inspection, there are two simple roots for the numerator, namely $a_1=0$ and $a_2=-11$ and so the solution of $$99 a^2-11979 a+11979=0$$ provides the other two $a_3,a_4$.
A: Since $a+b= 10c$ and $c+d=10a$ (Vieta's Formula ) ...
A: If we have a monic polynomial $x^2 + Ax + B$ for some numbers $A,B$, and this polynomial has roots $r_1, r_2$, then $A$ and $B$ are deeply related to $r_1$ and $r_2$. In particular, we must have
$$ x^2 + Ax + B = (x - r_1)(x-r_2) = x^2 - (r_1 + r_2)x + r_1r_2,$$
so that $-(r_1 + r_2) = A$ and $r_1r_2 = B$.
As an aside, this sort of pattern continues for higher degree polynomials. The coefficients of the polynomials are elementary symmetric polynomials in the roots.
So we are looking at the two polynomials
$$ \begin{align}
x^2 - 10cx - 11d \\
x^2 - 10ax - 11b
\end{align},$$
such that $a,b$ are the roots of the first and $c,d$ are the roots of the second. So we know that $a + b = 10c$ and $c + d = 10a$ by looking at the degree 1 coefficients, and similarly $ab = -11d$ and $cd = -11b$ from the constant coefficients.
By adding the identities of the degree $1$ coefficients, we see that $a + b + c + d = 10(a + c)$, or rather
$$ b + d = 9(a + c) \tag{1}.$$
So to understand $a + b + c + d$, it will suffice to understand either $b + d$ or $a + c$.
By multiplying the relations for the constant coefficients, we see that $abcd = 121bd$. We can cancel the $bd$ as long as $bd \neq 0$. Suppose that one of $b$ or $d$ is $0$, and without loss of generality suppose it is $b$. Then $b = 0$ is a root of the first polynomial $x^2 - 10cx - 11d$. This forces $d = 0$, so that $b = d = 0$. Since in the problem statement we are given that $b \neq d$, we see that $bd \neq 0$. So we may cancel off the $bd$, and we get that
$$ ac = 121 \tag{2}.$$
We go after either $a + c$. We know that $a$ is a root of the first polynomial and $c$ is a root of the second polynomial. So
$$\begin{align}
a^2 - 10ac - 11d = 0 \\
c^2 - 10ac - 11b = 0.
\end{align}$$
Notice we know that in the middle coefficients, we know $ac = 121$ by $(2)$. So adding these gives
$$ a^2 + c^2 - 20(121) - 11(b + d)=0.$$
We know that $11(b+d) = 99(a + c)$ by $(1)$ above, so this is
$$ a^2 + c^2 - 20(121) - 99(a + c)=0. \tag{3}$$
Notice that this is almost of the form $(a + c)^2 + \alpha (a + c) + \beta = 0$. We're off by $-2ac = -2(121)$. More explicitly, $(a + c)^2 - 2ac = (a+c)^2 - 2(121) = a^2 + c^2$, so that $(3)$ becomes
$$ (a+c)^2 - 99(a + c) - 22(121) = 0. \tag{4}$$
This is a quadratic in $a + c$. We know how to find the roots of
$$ x^2 - 99x - 22(121) = 0.$$
The quadratic formula tells us the roots are
$$ x = \frac{99 \pm 143}{2} = -22, 121$$
(and getting integer results here is a nice indication that this is the right path, and that they carefully designed the solution).
We must now ask whether both are actually possible. I had to think quite a bit on how to do this, and I find it a bit subtle. If $x = -22$, which is to say that $a + c = -22$, then in fact we have that
$$(a-c)^2 = (a+c)^2 - 4ac = 22^2 - 4(121) = 4 \cdot 11^2 - 4\cdot 11^2 = 0,$$
so that $a = c$. This contradicts the original claim that $a \neq c$, and so in fact $a + c = 121$.
Now that we know that $a + c = 121$, and we can relate $b + d$ to $a + c$ through $(1)$, we can get the answer, $a + b + c + d = 1210$. $\diamondsuit$
