How do you solve $(x^2+5x)/(x-3)>=0$ using a sign chart?

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How do you solve $(x^2+5x)/(x-3)>=0$ using a sign chart?

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Answer 1 · 2017-01-21T07:32:07

The answer is $x in [-5,0 ] uu ] 3, +oo [$

Explanation:

Let $f(x)=(x^2+5x)/(x-3)=(x(x+5))/(x-3)$

The domain of $f(x)$ is $D_f(x)=RR-{3}$

Let's do the sign chart

$color(white)(aaaa)$ $x$ $color(white)(aaaa)$ $-oo$ $color(white)(aaaa)$ $-5$ $color(white)(aaaaa)$ $0$ $color(white)(aaaaaaaaa)$ $3$ $color(white)(aaaaa)$ $+oo$

$color(white)(aaaa)$ $x+5$ $color(white)(aaaaa)$ $-$ $color(white)(aaaa)$ $+$ $color(white)(aaaaa)$ $+$ $color(white)(aaaa)$ $color(red)(∥)$ $color(white)(aa)$ $+$

$color(white)(aaaa)$ $x$ $color(white)(aaaaaaaa)$ $-$ $color(white)(aaaa)$ $-$ $color(white)(aaaaa)$ $+$ $color(white)(aaaa)$ $color(red)(∥)$ $color(white)(aa)$ $+$

$color(white)(aaaa)$ $x-3$ $color(white)(aaaaa)$ $-$ $color(white)(aaaa)$ $-$ $color(white)(aaaaa)$ $-$ $color(white)(aaaa)$ $color(red)(∥)$ $color(white)(aa)$ $+$

$color(white)(aaaa)$ $f(x)$ $color(white)(aaaaaa)$ $-$ $color(white)(aaaa)$ $+$ $color(white)(aaaaa)$ $-$ $color(white)(aaaa)$ $color(red)(∥)$ $color(white)(aa)$ $+$

Therefore,

$f(x)>=0$ , when $x in [-5,0 ] uu ] 3, +oo [$

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How do you solve $(x^2+5x)/(x-3)>=0$ using a sign chart?

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Explanation:

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How do you solve (x^2+5x)/(x-3)>=0(x^2+5x)/(x-3)>=0 using a sign chart?

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How do you solve $(x^2+5x)/(x-3)>=0$ using a sign chart?