Question

How do you solve `x^2+4x+3<=0`?

Answer 1

The solution is `x in [-3,-1]`

Explanation:

Let's factorise the inequality

`x^2+4x+3<=0`

`(x+1)(x+3)<=0`

Let, `f(x)=(x+1)(x+3)`

Now, we can build the sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `-3` `color(white)(aaaa)` `-1` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `x+3` `color(white)(aaaaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `x+1` `color(white)(aaaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+`

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

Therefore,

`f(x)<=0` when `x in [-3,-1]`