How do solve $(x-4)/(x^2+2x)<=0$ and write the answer as a inequality and interval notation?

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How do solve $(x-4)/(x^2+2x)<=0$ and write the answer as a inequality and interval notation?

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Answer 1 · 2016-12-26T14:23:36

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

Explanation:

Let's rewrite the inequality as

$(x-4)/(x^2+2x)=(x-4)/(x(x+2))$

And

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

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

Now, we can make the sign chart

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

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

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

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

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

Therefore,

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

or, $x<-2$ or $0< x <=4$

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How do solve $(x-4)/(x^2+2x)<=0$ and write the answer as a inequality and interval notation?

1 Answer

Explanation:

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How do solve (x-4)/(x^2+2x)<=0(x-4)/(x^2+2x)<=0 and write the answer as a inequality and interval notation?

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Explanation:

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How do solve $(x-4)/(x^2+2x)<=0$ and write the answer as a inequality and interval notation?