Question

How do you solve `(x+6)/(x^2+6x+8)>=0`?

Answer 1

The answer is `x in [-6, -4[uu ] -2,+oo [`

Explanation:

Let 's factorise the denominator

`x^2+6x+8=(x+4)(x+2)`

Let `f(x)=(x+6)/(x^2+6x+8)=(x+6)/((x+4)(x+2))`

The domain of `f(x)` is `D_f(x)=RR-{-2,-4}`

We solve the inequality with a sign chart

`color(white)(aaaa)` `x` `color(white)(aaaaaa)` `-oo` `color(white)(aaaa)` `-6` `color(white)(aaaa)` `-4` `color(white)(aaaaaa)` `-2` `color(white)(aaaaaa)` `+oo`

`color(white)(aaaa)` `x+6` `color(white)(aaaaaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aa)` `color(red)(∥)` `color(white)(aa)` `+` `color(white)(aa)` `color(red)(∥)` `color(white)(aaa)` `+`

`color(white)(aaaa)` `x+4` `color(white)(aaaaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aa)` `color(red)(∥)` `color(white)(aa)` `+` `color(white)(aa)` `color(red)(∥)` `color(white)(aaa)` `+`

`color(white)(aaaa)` `x+2` `color(white)(aaaaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aa)` `color(red)(∥)` `color(white)(aa)` `-` `color(white)(aa)` `color(red)(∥)` `color(white)(aaa)` `+`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aa)` `color(red)(∥)` `color(white)(aa)` `-` `color(white)(aa)` `color(red)(∥)` `color(white)(aaa)` `+`

Therefore,

`f(x)>=0`, when `x in [-6, -4[uu ] -2,+oo [`