<s>[INST] $a,b$ 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
 [/INST] 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 $a+b+c+d = 3(a+c) = 3\cdot 32 = 96\Rightarrow (a+b+c+d) = 96$
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