<s>[INST] 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 $a+b+c+d=?$
 [/INST] 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 $a_3,a_4$.
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