How do you solve $(x-4)/(x^2+2x)<=0$ using a sign chart?

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How do you solve $(x-4)/(x^2+2x)<=0$ using a sign chart?

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Answer 1 · 2016-12-14T10:33:31

The answer is $x in ] -oo,-2 [ uu ] 0,4]$

Explanation:

The denominator is

$x^2+2x=x(x+2)$

Let $f(x)=(x-4)/(x(x+2))$

The domain of $f(x)$ is $D_f(x)=RR-{0,-2}$

Now, we can do the sign chart

$color(white)(aaaa)$ $x$ $color(white)(aaaa)$ $-oo$ $color(white)(aaaa)$ $-2$ $color(white)(aaaa)$ $0$ $color(white)(aaaa)$ $4$ $color(white)(aaaa)$ $+oo$

$color(white)(aaaa)$ $x-2$ $color(white)(aaaaa)$ $-$ $color(white)(aa)$ $∥$ $color(white)(a)$ $+$ $color(white)(aa)$ $+$ $color(white)(aaa)$ $+$

$color(white)(aaaa)$ $x$ $color(white)(aaaaaaaaa)$ $-$ $color(white)(aa)$ $∥$ $color(white)()$ $-$ $color(white)()$ $∥$ $color(white)(a)$ $+$ $color(white)(aaa)$ $+$

$color(white)(aaaa)$ $x-4$ $color(white)(aaaaa)$ $-$ $color(white)(aa)$ $∥$ $color(white)()$ $-$ $color(white)()$ $∥$ $color(white)(a)$ $-$ $color(white)(aaa)$ $+$

$color(white)(aaaa)$ $f(x)$ $color(white)(aaaaaa)$ $-$ $color(white)(aa)$ $∥$ $color(white)()$ $+$ $color(white)()$ $∥$ $color(white)(a)$ $-$ $color(white)(aaa)$ $+$

So,
$f(x)<=0$ , when $x in ] -oo,-2 [ uu ] 0,4]$

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How do you solve $(x-4)/(x^2+2x)<=0$ using a sign chart?

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Explanation:

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