IIT JEE , AIEEE Forum - Mathematics , Trignometry

me

hey dude the question is dam good.

ok here is a solution------
x^2+19x+88=(x+11)(x+8)

now for it to be a square either x+11=x+8 or x+11 & x+8 both has to be square

case 1--- we get 11=8 absurd

case2 --- job at hand is simply to find 2 nos whith squares having difference of 3(as 11- 7 =3)

this possible only with (1,2) &(-1,-2)

hence required values of x = -7 &-12

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