How do you solve $\frac{{(x + 1)}^{2}}{x^{2} + 2 x - 3} \leq 0$ $(x+1)^2/(x^2+2x-3)<=0$ using a sign chart?

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Answer 1 · 2016-12-16T17:49:54

The answer is $x \in] - 3, 1 [$ $x in ] -3,1 [$

Let's factorise the denominator

$x^{2} + 2 x - 3 = (x - 1) (x + 3)$ $x^2+2x-3=(x-1)(x+3)$

Let $f (x) = \frac{{(x + 1)}^{2}}{(x - 1) (x + 3)}$ $f(x)=(x+1)^2/((x-1)(x+3))$

The domain of $f (x)$ $f(x)$ is $D_f(x)=RR-{-3,1}$

The numerator $(x+1)^2>0, AAx in D_f(x)$

Now we can do the sign chart

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

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

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

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

Therefore,

$f(x)<=0$ when $x in ] -3,1 [$

graph{y-(x+1)^2/(x^2+2x-3)=0 [-12.66, 12.65, -6.33, 6.33]}

How do you solve (x+1)^2/(x^2+2x-3)<=0(x+1)2x2+2x−3≤0(x+1)^2/(x^2+2x-3)<=0 using a sign chart?