How do you solve $(x^2-x-2)/(x+3)<0$ ?

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How do you solve $(x^2-x-2)/(x+3)<0$ ?

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Answer 1 · 2017-03-08T10:55:22

The solution is $x in ]-oo, -3[ uu ]-1,2[$

Explanation:

We solve this inequality with a sign chart

Let's factorise the numerator

$x^2-x-2=(x+1)(x-2)$

The inequality is

$(x^2-x-2)/(x+3)=((x+1)(x-2))/(x+3)<0$

Let $f(x)=((x+1)(x-2))/(x+3)$

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

We can build the sign chart

$color(white)(aaaa)$ $x$ $color(white)(aaaa)$ $-oo$ $color(white)(aaaaa)$ $-3$ $color(white)(aaaaaa)$ $-1$ $color(white)(aaaa)$ $2$ $color(white)(aaaaaa)$ $+oo$

$color(white)(aaaa)$ $x+3$ $color(white)(aaaa)$ $-$ $color(white)(aaaa)$ $||$ $color(white)(aaaa)$ $+$ $color(white)(aaaa)$ $+$ $color(white)(aaaa)$ $+$

$color(white)(aaaa)$ $x+1$ $color(white)(aaaa)$ $-$ $color(white)(aaaa)$ $||$ $color(white)(aaaa)$ $-$ $color(white)(aaaa)$ $+$ $color(white)(aaaa)$ $+$

$color(white)(aaaa)$ $x-2$ $color(white)(aaaa)$ $-$ $color(white)(aaaa)$ $||$ $color(white)(aaaa)$ $-$ $color(white)(aaaa)$ $-$ $color(white)(aaaa)$ $+$

$color(white)(aaaa)$ $f(x)$ $color(white)(aaaaa)$ $-$ $color(white)(aaaa)$ $||$ $color(white)(aaaa)$ $+$ $color(white)(aaaa)$ $-$ $color(white)(aaaa)$ $+$

Therefore,

$f(x)<0$ when $x in ]-oo, -3[ uu ]-1,2[$

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How do you solve $(x^2-x-2)/(x+3)<0$ ?

1 Answer

Explanation:

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How do you solve $(x^2-x-2)/(x+3)<0$ ?