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Let the 2 roots be (10a+k) and (10b+k) here k is unit digit of both roots so sum of roots = -b/a product of roots = c/a (10(a+b) +2k) = -b/a and (10a+k)*(10b+k) = c/a k = -b/2a -5(a+b) which on substitution gives : (5(a-b) - b/2a)*(5(b-a) + b/2a) = c/a (b/2a - 5(a-b))*(5(a-b) + b/2a) = c/a b2/4a2 - 25(a-b)2 = c/a (b2 - 4ac)/a2 = 100(a-b)2 but given that (b2 - 4ac)/a2 is not multiple of 100 [ as a,b are intezers, a-b
so (a-b)2<1
i.e., possible value of : b/2a =1/2
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let the roots are 10p and 10q.
now,
sum of roots: 10(p+q)= -b/a.
product of roots is 100pq= c/a
value of d/a^2 = 100(p-q)^2.
since d/a^2 is not a multiple of 100, hence (p-q) is not integer.
thus p and q are not integers.
but for roots being integers p and q should be multiple of 1/ 2 or 1/ 5.
let both are multiple of 1/2
then both roots must have 0 or 5 at its unit place.
thus the possible value of unit digit of b/2a is 0 or 5.
hence the correct option is b.
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