Integers and Inequality

Since absolute values are always non-negative, we know that the numerator will always be non-negative. Therefore, for the expression to be negative, the denominator must be negative and the numerator must be strictly positive. Factoring the denominator gives us

$x^2-36x = x(x-36)$

The zeroes of this expression are 0 and 36.

If $x>36$ , both factors $x$ and $x-36$ are positive, and their product is positive.

If $x<0$ , both factors are negative, and their product is positive.

If $0<x<36$ , the factor $x$ is positive and the factor $x-36$ is negative, and their product is negative.

So we've narrowed down the possible solutions to the integers from 1 to 35 inclusive. Now we just need to determine if any of these integers make the numerator zero. Solving the equation $x^2-81=0$ for $x$ , we find that the numerator is zero when $x=\pm 9$ . Therefore, the solutions are all of the integers from 1 to 35 inclusive except 9, so there are 34 integer values of $x$ that satisfy the inequality.

If x is equal to any negative integer then result will be positive.

If x=0 we get undefined result. If x=9 then results is 0.And also x=36

we get undefined results so we have numbers from 1 to 35 except 9.

if x=9 then x2-81=0 and if x=36 then x2-36x=0. So result is 34.