Question

How do you solve `(x^2-x-2)/(x^2-4x+3)>0`?

Answer 1

`x in (-oo, -1) uu (1, 2) uu (3, oo)`

Explanation:

`(x^2-x-2)/(x^2-4x+3) = ((x-2)(x+1))/((x-1)(x-3))`

So the numerator changes sign at `x=-1` and `x=2` while the denominator changes sign at `x=1` and `x=3`

Each change of sign (numerator or denominator) changes the sign
of the quotient.

So the sign reverses each time we move to the next interval in the sequence:

`(-oo, -1), (-1, 1), (1, 2), (2, 3), (3, oo)`

For large positive values of `x`, all of the linear factors are positive, so `(x^2 - x - 2)/(x^2-4x+3) > 0`.

So the signs of the quotient in the `5` intervals listed are:

`+ - + - +`

So the quotient is positive in `(-oo, -1) uu (1, 2) uu (3, oo)`

graph{(x^2-x-2)/(x^2-4x+3) [-10, 10, -5, 5]}