Question

How do solve `(x^2+5x)/(x-3)>=0` and write the answer as a inequality and interval notation?

Answer 1

The answer is `x in [-5,0 ] uu ]3, +oo[`

Explanation:

Let `f(x)=(x^2+5x)/(x-3)=(x(x+5))/(x-3)`

The domain of `f(x)` is `D_(f(x))=RR-{3}`

Let's do a sign chart to solve this inequality

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `-5` `color(white)(aaaa)` `0` `color(white)(aaaa)` `3` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `x+5` `color(white)(aaaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaa)` `+` `color(white)(aaa)` `+`

`color(white)(aaaa)` `x` `color(white)(aaaaaaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aa)` `+` `color(white)(aaa)` `+`

`color(white)(aaaa)` `x-3` `color(white)(aaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaa)` `-` `color(white)(aaa)` `+`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaa)` `-` `color(white)(aaa)` `+`

Therefore `f(x)>=0` when `x in [-5,0 ] uu ]3, +oo[`

graph{(x^2+5x)/(x-3) [-72.6, 75.47, -19.23, 54.86]}