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 $c=2a-d$.
Best Answer
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 \ \ $ $p =\pm \dfrac{5\sqrt{5}}{2}$.