if the roots of the equation x^2-10cx-11d=0 are a,b and those of x^2-10ax-11b=0 are c,d then find the value of a+b+c+d(a,b,c,d real distinct numbers)?
(ans=1210)
Boy,your answer is surely wrong
$\displaystyle a+b=10c$
$\displaystyle c+d=10a$
$\displaystyle a+b+c+d=10(a+c)$
$\displaystyle b+d=9(a+c).$Also we have $\displaystyle a+b-c-d=10c-10a$.Thus, $\displaystyle b-d=11(c-a)$
Since $\displaystyle a$ is root of $\displaystyle x^2-10cx-11d=0$,we have $\displaystyle a^2-10ac-11d=0$........(1)
Since $\displaystyle c$is root of $\displaystyle x^2-10ax-11b=0$,we have $\displaystyle c^2-10ac-11b=0$.........(2)
Subtracting (1) and (2)
$\displaystyle c^2-a^2=11(b-d)$
$\displaystyle (a+c)(c-a)=121(c-a)$
$\displaystyle (a+c)=121$
$\displaystyle a+b+c+d=10(a+c)=1210$