[administration]

Q. How many real “ r” are there such that the roots of x² + rx + 6r = 0 are both integers?
(a) 5 (b) 9 (c) 10 (d) 18 (e) none of these

[Wonder Boy]

Let the roots of equation be u & v.

u + v =-r
uv = 6r
1/u +1/v = -1/6 –> u= -6v/(6+v) = [-6(6+v) +36]/(6+v)

u = -6 +36/(6+v)

For both the roots to be integers
(6+v) should be the factor of 36
values that v can take as
{0,3,6,12,30, -2,-3,-4,-5,-7,-8,-9,-10,-12,-15,-18,-24,-42}

and corresponding values of a .

so, answer is 18

[Author]

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