Question

How do you solve `(x^2-4)/(x^2-4x-5) > 0` ?

Answer 1

`x in (-oo, -2) uu (-1, 2) uu (5, oo)`

Explanation:

To solve:

`(x^2-4)/(x^2-4x-5) > 0`

Multiply both sides by the square of the denominator to get:

`0 < (x^2-4)(x^2-4x-5) = (x-2)(x+2)(x-5)(x+1)`

Each of the zeros of the right hand side is of multiplicity `1`, so this product changes sign at `x=-2`, `x = -1`, `x=2` and `x=5`.

It has constant sign in all of the intervals between them.

In addition, note that the sign of the leading coefficient is positive, so the product is positive for large positive values of `x`.

Hence it is positive in `(-oo, -2) uu (-1, 2) uu (5, oo)`

Here's a graph of the original function:
graph{(x^2-4)/(x^2-4x-5) [-10, 10, -15, 15]}

Here's a graph of the function as multiplied by the square of the denominator:
graph{(x^2-4)(x^2-4x-5) [-10, 10, -70, 55]}

Notice that they are both positive or negative for the same values of `x`.