$a$ 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 $c=2a-d$.
 A: Let the two polynomials be $$
p(x) = x^2 - 2cx -5d \\
q(x) = x^2 - 2ax - 5b
$$
You also know that $$
p(x) = (x-a)(x-b) = x^2 - (a+b)x + ab \\
q(x) = (x-c)(x-d) = x^2 - (c+d)x + cd
$$
You might try playing around with these two forms.  For example, you can take the product of the polynomials and equate the coefficients for each power of $x$.
A: Vieta's formulas $\Rightarrow \ \ $ $a+b=2c$, $c+d=2a$ $\ \ \Rightarrow \ \ $ $a+b+c+d=2(a+c)$ $\ \Rightarrow \ $ $a+c=b+d$.
Denote
$$m = \dfrac{a+c}{2}=\dfrac{b+d}{2}, \qquad  p=\dfrac{c-a}{2}\color{gray}{=m-a=c-m}.$$
Then
$$
\left\{
\begin{array}{r}
a = m-p, \quad b=m+3p, \\
c = m+p, \quad d=m-3p.
\end{array}
\right.
$$
Vieta's formulas $\Rightarrow \ $ $ab=-5d$, $\ \ $ $cd=-5b$ $\ \ \Rightarrow$
$$
\left\{
\begin{array}{r}
(m-p)(m+3p)=-5m+15p, \\
(m+p)(m-3p)=-5m-15p;
\end{array}
\right.
$$
$$
\left\{
\begin{array}{r}
m^2+2mp-3p^2=-5m+15p, \\
m^2-2mp-3p^2=-5m-15p;
\end{array}
\right.
$$ 
$$
\left\{
\begin{array}{c}
m^2-3p^2=-5m, \\
2mp=15p.
\end{array}
\right.
$$ 
Since $a,b,c,d$ are distinct, then $p\ne 0$, then $2m=15$, then 
$$\color{#660011}{\Large{a+b+c+d=4m=30}}.$$
Note:
$3p^2=m^2+5m=\dfrac{375}{4}$ $\ \ \Rightarrow \ \ $ $p =\pm \dfrac{5\sqrt{5}}{2}$.
A: $\bf{My\; Solution::}$ 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 $a+b+c+d = 2(a+c) = 2\cdot 15 = 30\Rightarrow \boxed{\boxed{a+b+c+d = 30}}$
A: For a hint, try plugging in $a$ and $b$ into your first equation since you know that they make the equality true.  That will get you two equations.  You can do the same thing with $c$ and $d$ in the second equation.
