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

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Answer 1 · 2017-01-13T17:52:52

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

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 $x in [-5, 0]$

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