<s>[INST] If $x^2 - 10ax - 11b = 0$ has roots $c$ and $d$ and $x^2 - 10cx - 11d = 0$ has roots $a$ and $b$, then find $a + b + c + d$? If $x^2 - 10ax - 11b = 0$ have roots $c$ and $d$ and $x^2 - 10cx - 11d = 0$ have roots $a$ and $b$, then find $a + b + c + d$?
My attempt-
Using Vieta's formula, we get
$$ c+d=10a, a+b=10c $$
Because sum of roots $=\frac{-b}{a}$
Also using Vieta's formula we get
$$ cd=-11b, ab=-11d $$
Because product of roots $=\frac{c}{a}$
So now we need to solve these equations-
$$ c+d=10a, a+b=10c, cd=-11b, ab=-11d $$
So I solved it like this-
In the above line let each 4 equation be eq $1,2,3,4$.
Dividing eq $1$ by $2$ and cancelled $10$ in RHS because its both in numerator and denominator,
$$ \frac{c+d}{a+b}=\frac{a}{c} $$
Using componendo alone we get
$$ \frac{c+d+a+b}{a+b}=\frac{a+c}{c} $$
Putting all the values using eq $1$ and $2$,
$$ \frac{10a+10c}{ac}=\frac{a+c}{c} $$
Solving this we get $a=10$
Now doing mischiefs(some changes) with eq $1$,
$$d=10a-c  =>d=100-c  $$
Let the above equation be equation $5$.
Similarly doing some mischiefs with equation $2$ also
$$ b=10c-10 $$
Let above equation be equation $6$.
Now using equation $4$,
$$ ab=-11d $$
We know $a=10$ and using equation $5$ and $6$.
$$ 10(10c-10)=-11(100-c) $$
$$ 100c-100=-1100+11c $$
$$ 89c+1000=0 $$
$$ c=\frac{-1000}{89} $$
Now see adding equation $1$ and $2$
$$a+b+c+d=10a+10c$$
Thus we need to only find now $10(a+c)$.
$$ 10(a+c) $$
$$ 10(10+\frac{-1000}{89}) $$
But wait this is the wrong answer according to my book. What have I done wrong?
Here is the correct solution according to a website-
https://www.sarthaks.com/?qa=blob&qa_blobid=286792613098888701
Please help me with this question. Apologies for asking a homework question. Apologies for something missing in the question. Thanks for your time.
 [/INST] In your approach, after doing componendo alone, you substitute $a+b = ac$, which should $10c$ and it shouldn't give you anything new. Instead, we can do as follows which is straightforward calculation.
$d = 10 a - c$ and $b = 10c - a$
$(ab)(cd) = (-11b)(-11d) = 121bd$
If $b=0$, then $d = 0 = a = c$ and same for $d = 0$, $a=0$ and $c=0$ cases
Otherwise, $ac = 121$
$ab = 10ac - a^2 = -11d = -110a + 11 c$
$a^2 -110a +11\frac{121}{a} - 1210 = 0$
$a^3 - 10*11a^2 - 10*11^2a + 11^3 = 0$
$(a^3+11^3) - 110a(a + 11) = 0$
$(a+11)(a^2 - 121a + 11^2) = 0$
$a$ has 3 solutions and find solution set for each case ($c = \dfrac{121}{a}$, $d = 10 a - c$ and $b = 10c - a$).
</s>