Question

How do you solve `(x^2+5x)/(x-3)>=0` using a sign chart?

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 the sign chart

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

`color(white)(aaaa)` `x+5` `color(white)(aaaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaaa)` `+` `color(white)(aaaa)` `color(red)(∥)` `color(white)(aa)` `+`

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

`color(white)(aaaa)` `x-3` `color(white)(aaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaaa)` `-` `color(white)(aaaa)` `color(red)(∥)` `color(white)(aa)` `+`

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

Therefore,

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