How many integers satisfy the inequality |10(x+1)/x^2+2x+3|=1?
Solution) first thing thats not an inequality,
and second thing its very easy if thats the question.
the LHS = |10/x^2 + 10/x + 2x + 3| = 1
10/x^2 will also be an integer only if x^2<10.....i.e. x<3.
hence x= 0,+1,-1,+2,-2,+3,-3.
but the only value that holds the equality is -1.
hence the no. of INTEGERS is only 1.