<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] 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.
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