Question

How do solve `(x-4)/(x^2+2x)<=0` and write the answer as a inequality and interval notation?

Answer 1

The answer is `x<-2` or `0< x <=4`
`=x in ] -oo,-2 [ uu ] 0, 4]`

Explanation:

Let's rewrite the inequality as

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

And

Let `f(x)=(x-4)/(x(x+2))`

The domain of `f(x)` is `D_f(x)=RR-{-2,0}`

Now, we can make the sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `-2` `color(white)(aaaaaa)` `0` `color(white)(aaaaaa)` `4` `color(white)(aaaa)` `+oo`

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

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

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

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

Therefore,

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

or, `x<-2` or `0< x <=4`