Solving Rational Inequalities: How to Find the Solution Set?

[Nelo]

Homework Statement

x^2 +3x +2
__________ < 0
x^2-9

Homework Equations

The Attempt at a Solution

I factored the top to be

(x+2) (x+1)
______________ < 0
(x-3)(x+3)

Couldnt cancel anything out, So now i don't know what to do. I missed this lesson in class and i don't really understand wtf I am supposed to do. Inequalities usually have 1-3 cases in which we solve for x and create a therefore statement in, however, how am i supposed to solve this?

In the book it just shows intervals, but is it possible to solve through 3 cases as wel? ie)
x < -3 , x > 3 (since those are the vertical asymptotes)

 

[Nelo]

anyone?

 

[eumyang]

I use sign charts to solve rational inequalities. See http://www.purplemath.com/modules/ineqsolv3.htm" for an example.

P.S. Why do you continue to bump your thread before waiting 24 hours? It's against the rules, so I've reported you to the mods.

 

[Nelo]

Cuz i dun got 24 hours to wait for a small response. nt tho kid gl

 

[symbolipoint]

(x+2) (x+1)
______________ < 0
(x-3)(x+3)

The critical values of x of {+3, -3} at which the rational expression is undefined, and the critical values of x of {-2, -1} at which the rational expression is zero, contain the the values which break the Real Numbers into intervals which you can check.

You then can check any value within EACH of the intervals of the Real Numbers to determine if the value makes the inequality true or false. The intervals to check are obviously (-∞, -3), (-3, -2), (-2, -1), (-1, +3), and {+3, ∞).

 

[Mark44]

P.S. Why do you continue to bump your thread before waiting 24 hours? It's against the rules, so I've reported you to the mods.

Report noted, and acted on.

 

[Evo]

Cuz i dun got 24 hours to wait for a small response. nt tho kid gl

Knock off the childish text speak and type in normal English.

 

[Evo]

Homework Statement

x^2 +3x +2
__________ < 0
x^2-9

Homework Equations

The Attempt at a Solution

I factored the top to be

(x+2) (x+1)
______________ < 0
(x-3)(x+3)

Couldnt cancel anything out, So now i don't know what to do. I missed this lesson in class and i don't really understand wtf I am supposed to do. Inequalities usually have 1-3 cases in which we solve for x and create a therefore statement in, however, how am i supposed to solve this?

In the book it just shows intervals, but is it possible to solve through 3 cases as wel? ie)
x < -3 , x > 3 (since those are the vertical asymptotes)

Copy.