If roots of the equation x 2 −10cx−11d=0 are a, b and those of x 2 −10ax−11b=0 are c, d, then the value of a + b + c + d is (a, b, c and d are distinct numbers)

Answer 1

x^2-10c-11d=0

(x-5c)^2 = 11d +25c^2

x-5c = +/-sqr(11d +25c^2)

x = 5c +sqr(11d+25c^2) = a and 5c-sqr(11d+25c^2) = b

x^2 -10a -11b = 0

x = 5a +sqr(11b+25a^2) =c, 5a+sqr(11b +25a^2) = d

a+b+c+d = 10c +10a

the possibly irrational values cancel out

let a and b be values that make 11b +25a^2 a perfect square and then d and c are integers, a rational number.