<s>[INST] $a$ and $b$ are the roots of quadratic equation $x^2 -2cx-5d=0$ and $c$ and $d$ are the roots of quadratic equation $x^2 -2ax-5b=0 $ Let $a,\,b,\,c,\,d$ be distinct real numbers and $a$ and $b$ are the roots of quadratic equation $x^2 -2cx-5d=0$ and $c$ and $d$ are the roots of quadratic equation $x^2 -2ax-5b=0$. Then find the value of $a+b+c+d$.
I  could only get $2$ equations that 
$a=2c-b$ and $c=2a-d$.
 [/INST] $\bf{My\; Solution::}$ Given $a\;,b$ are the roots of the equation $x^2-2cx-5d=0$ So 
$\displaystyle a+b=2c............................(1)\;\;\;\;\;\; ab = -5d.....................(2)$
similarly $c\;,d$ are the roots of the equation $x^2-2ax-5b=0$ So
$\displaystyle c+d=2a............................(3)\;\;\;\;\;\; cd=-5b.....................(4)$
So $a+b+c+d = 2(a+c)............(5)$
Now $\displaystyle \frac{a+b}{c+d}=\frac{2c}{2a}=\frac{c}{a}\Rightarrow a^2+ab=c^2+cd\Rightarrow (a^2-c^2)=(cd-ab)=-5(b-d)$
So $\displaystyle (a+c)\cdot (a-c)=-5(b-d)=-5\left\{(2c-a)-(2a-c)\right\}=-15\left\{c-a\right\}=15(a-c)$
So $\displaystyle (a+c)\cdot (a-c)-15(a-c)=0\Rightarrow (a-c) = 0$ or $(a+c) = 15$
Now $a\neq c\;,$ bcz $a,b,c,d$ are distinct real no.
So $a+c-15=0\Rightarrow a+c = 15$. put into eqn....$(5)$
We get $a+b+c+d = 2(a+c) = 2\cdot 15 = 30\Rightarrow \boxed{\boxed{a+b+c+d = 30}}$
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