$a,b$ are roots of $x^2-3cx-8d = 0$ and $c,d$ are roots of $x^2-3ax-8b = 0$. Then $a+b+c+d =$ (1) If $a,b$ are the roots of the equation $x^2-10cx-11d=0$ and $c,d$ are the roots of the equation 
$x^2-10ax-11b=0$. Then the value of $\displaystyle \sqrt{\frac{a+b+c+d}{10}}=,$ where $a,b,c,d$ are distinct real numbers.
(2) If $a,b,c,d$ are distinct real no. such that $a,b$ are the roots of the equation $x^2-3cx-8d = 0$
and $c,d$ are the roots of the equation $x^2-3ax-8b = 0$. Then $a+b+c+d = $
$\bf{My\; Try}::$(1) Using vieta formula
$a+b=10c......................(1)$ and $ab=-11d......................(2)$
$c+d=10a......................(3)$ and $cd=-11b......................(4)$
Now $a+b+c+d=10(a+c)..........................................(5)$
and $abcd=121bd\Rightarrow bd(ab-121)=0\Rightarrow bd=0$ or $ab=121$
Now I did not understand how can i calculate $a$ and $c$
Help Required
Thanks
 A: The answer for (1) is $11$.
$$abcd=121bd\Rightarrow bd(ac-121)=0\Rightarrow bd=0\ \text{or}\ ac=121.$$
(Note that you have a mistake here too.)
1) The $bd=0$ case : If $b=0$, we have $x(x-10a)=0$. This leads that $c=0$ or $d=0$. This is a contradiction. The $d=0$ case also leads a contradiction.
2) The $ac=121$ case : We have $$c=\frac{121}{a}, b=\frac{1210}{a}-a, d=10a-\frac{121}{a}.$$ Hence, we have
$$1210-a^2-11\left(10a-\frac{121}{a}\right)=0$$
$$\Rightarrow a^3-110a^2-1210a+121\times 11=0$$
$$\Rightarrow a=11, \frac{11(11\pm 3\sqrt{13})}{2}.$$
If $a=11$, then $c=11$, which is a contradiction. Hence, we have
$$(a,c)=\left(\frac{11(11\pm 3\sqrt{13})}{2},\frac{11(11\mp 3\sqrt{13})}{2}\right).$$
Hence, we have
$$\sqrt{\frac{a+b+c+d}{10}}=\sqrt{a+c}=\sqrt{121}=11.$$
I think you can get an answer for (2) in the same way as above.
A: Is anything more given? If not there are 4 answes, and two are integers.
Here is one answer: a=b=c=d=0
The other answer: a=c =-11, b=d=-99
There are two other irrational solutioms for $a$,$b$, $c$,$d$ with $z=11$
Can you provide more info? I will edit this answer  based on what you tell me
A: Thanks mathlove.
My Solution for (2) one::
Given $a,b$ are the roots of $x^2-3cx-8d=0.$ So $a+b=3c.........(1)$ and $ab=-8d........(2)$
and $c,d$ are the roots of $x^2-3ax-8b=0.$ So $c+d=3a..........(3)$ and $cd=-8b..........(4)$
So $a+b+c+d = 3(a+c)......................(5)$
Now $\displaystyle \frac{a+b}{c+d} = \frac{3c}{3a}=\frac{c}{a}\Rightarrow a^2+ab=c^2+cd\Rightarrow a^2-8d=c^2-8b$
So $(a^2-c^2)=-8(b-d)\Rightarrow (a+c)\cdot(a-c)=-8(b-d).................(6)$
Now $(1)-(3)$, we get $a+b-c-d=3c-3a\Rightarrow (b-d) = 4(c-a)=-4(a-c)$
Now put $(b-d) = -4(a-c)$ in eqn... $(6)$, we get $(a+c)\cdot(a-c)=32(a-c)$
So $(a-c)\cdot(a+c-32) = 0$, Now $a\neq c$, bcz $a,b,c,d$ are distinct real no.
So $a+c=32$, Put into eqn...$(5)$
We get $a+b+c+d = 3(a+c) = 3\cdot 32 = 96\Rightarrow (a+b+c+d) = 96$
