How do you solve: (x^2-9)/(x^2-1) < 0?

1 Answer
Oct 20, 2015

x in (-3, -1) uu (1, 3)

Explanation:

Your inequality looks like this

(x^2 - 9)/(x^2 - 1) < 0

Right from the start, you know that any solution set that you might come up with cannot include the values of x that will make the denominator equal to zero.

More specifically, you need to have

x^2 - 1 != 0 implies x != +- 1

Now, in order for this inequality to be true, you need to have

x^2 - 9 < 0" " and " "x^2 - 1 > 0

or

x^2 - 9 >0" " and " "x^2 - 1 < 0

For the fist set of conditions to be true, you need to have

{(x^2 - 9 < 0 implies x < +- 3 implies x in (-3, 3)), (x^2 - 1 > 0 implies x > +- 1 implies x in (-oo, -1) uu (1, + oo)) :}

This means that you need x in (-3, -1) uu (1, 3).

For the second set of conditions, you need to have

{(x^2 - 9 > 0 implies x > +- 3 implies x in (-oo, -3) uu (3, + oo)), (x^2 - 1 < 0 implies x < +- 1 implies x in (-1, 1)) :}

This time, those two intervals will not produce a valid solution set, or x in O/.

The only option left to you is x in (-3, -1) uu (1,3). The values of x that belong to this interval will make the numerator negative and the denominator positive, which in turn will make the fraction negative.

graph{(x^2 - 9)/(x^2 - 1) [-18.02, 18.01, -9.01, 9.01]}