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				   The 2 roots of ax^2+bx+c=0 are both 2 digit integers with the same units digit, but D/a^2 (D=Discriminant) is not a multiple of 100. Which is a possible value of the unit digit of b/2a?

a)4, b)5, c)6, d)7

7 years ago


Answers : (2)


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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Naga Ramesh

IIT Kgp - 2005 batch

7 years ago

let the roots are 10p and 10q.


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.

7 years ago

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