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(C) Given that $a, b$ are roots of $x^2-5cx-6d=0$,we have:$a+b=5c$ $(1)$$ab=-6d$ $(2)$Given that $c, d$ are roots of $x^2-5ax-6b=0$,we have:$c+d=5a$ $

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$144$

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(C) Given that $a, b$ are roots of $x^2-5cx-6d=0$,we have:$a+b=5c$ $(1)$$ab=-6d$ $(2)$Given that $c, d$ are roots of $x^2-5ax-6b=0$,we have:$c+d=5a$ $(3)$$cd=-6b$ $(4)$Subtracting $(3)$ from $(1)$:$(a+b)-(c+d) = 5c-5a$$(a-c)+(b-d) = -5(a-c)$$b-d = -6(a-c) = 6(c-a)$ $(5)$Adding $(1)$ and $(3)$:$(a+b)+(c+d) = 5c+5a$$(b+d) = 4(a+c)$ $(6)$From $(2)$ and $(4)$,$ab-cd = -6d+6b = 6(b-d)$.Substituting $(5)$ into this: $ab-cd = 6(6(c-a)) = 36(c-a)$.Also,$ab-cd = a(5c-a) - c(5a-c) = 5ac - a^2 - 5ac + c^2 = c^2-a^2 = (c-a)(c+a)$.Since $a, b, c, d$ are distinct,$c-a 
eq 0$,so $c+a = 36$.Substituting this into $(6)$: $b+d = 4(36) = 144$.

Let $a, b, c, d$ be distinct real numbers such that $a, b$ are roots of $x^2-5cx-6d=0$,and $c, d$ are roots of $x^2-5ax-6b=0$. Then $b+d$ is

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