Question

How do you solve `\frac { x ^ { 2} + 3x + 2} { x ^ { 2} - 9} \leq 0`?

Answer 1

The solution is `x in (-3,-2] uu [-1,3)`

Explanation:

Let`s factorise the numerator and denominator

`(x^2+3x+2)/(x^2-9)<=0`

`((x+1)(x+2))/((x+3)(x-3))<=0`

Let

`f(x)=((x+1)(x+2))/((x+3)(x-3))`

We can build the sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaaa)` `-3` `color(white)(aaaa)` `-2` `color(white)(aaaaa)` `-1` `color(white)(aaaaaaaa)` `3` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `x+3` `color(white)(aaaa)` `-` `color(white)(aaaa)` `||` `color(white)(aaa)` `+` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+` `color(white)(aaaa)` `||` `color(white)(aa)` `+`

`color(white)(aaaa)` `x+2` `color(white)(aaaa)` `-` `color(white)(aaaa)` `||` `color(white)(aaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+` `color(white)(aaaa)` `||` `color(white)(aa)` `+`

`color(white)(aaaa)` `x+1` `color(white)(aaaa)` `-` `color(white)(aaaa)` `||` `color(white)(aaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `||` `color(white)(aa)` `+`

`color(white)(aaaa)` `x-3` `color(white)(aaaa)` `-` `color(white)(aaaa)` `||` `color(white)(aaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `||` `color(white)(aa)` `+`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaa)` `+` `color(white)(aaaa)` `||` `color(white)(aaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `-` `color(white)(aaaa)` `||` `color(white)(aa)` `+`

Therefore,

`f(x)<=0` when `x in (-3,-2] uu [-1,3)`