Question

How do you solve `x^2+5x<=0` using a sign chart?

Answer 1

The answer is `=x in [-5, 0]`

Explanation:

Let's factorise the equation

`x^2+5x=x(x+5)`

Let `f(x)=x(x+5)`

Let's do the sign chart

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

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

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

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

Therefore,

`f(x<=0)` when `x in [-5, 0]`