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

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Answer 1 · 2017-02-28T07:43:58

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

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 $x in [-3,-1]$

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