Quadratic equation.. not too difficult..

f(x) = (x - a) ( x-c) + 2(x-b)(x-d)=0

f(a) = 0 + 2 (a-b) (a-d) > 0 {bcoz a<b<c<d and so a-b < 0 and a-d < 0 }

similarily
f(b) = (b-a) (b-c) + 0 < 0 {bcoz b-a > 0 and b-c < 0}

Since f(a) and f(b) are of opposite signs, therefore, then f(x) = 0 will have atleast one real root b/w a and b

Similarily
f(c) = 0 + 2 (c-b) (c-d) < 0 {bcoz c-b > 0 and c-d < 0}

f(d) = (d-a) (d-c) + 0 > 0 {bcoz d-a >0 and d-c < 0}

Since f(c) and f(d) are of opp signs therefore, one root lie b/w c and d .

Hence the roots of the eqn f(x) = 0 are real and distinct.
hence (b) option.

For a more general discussion see http://www.goiit.com/posts/list/algebra-try-this-if-a-b-c-d-then-prove-that-for-any-real-the-977445.htm#1161137