# How do solve (x+1)^2/(x^2+2x-3)<=0 and write the answer as a inequality and interval notation?

Dec 13, 2016

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

#### Explanation:

Let's factorise the denominator

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

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

The domain of $f \left(x\right)$ is ${D}_{f} \left(x\right) = \mathbb{R} - \left\{- 3 , 1\right\}$

The numerator is $> 0 , \forall x \in {D}_{f} \left(x\right)$

Now, we can do the sign chart

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$- 3$$\textcolor{w h i t e}{a a a a}$$1$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x + 3$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$x - 1$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a}$$+$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$

So,

$f \left(x\right) \ge 0$ when  x in ] -3,1 [  or $- 3 < x < 1$

Let $g r a p h \left\{y = {\left(x + 1\right)}^{2} / \left(\left(x - 1\right) \left(x + 3\right)\right) \left[- 12.66 , 12.65 , - 6.33 , 6.33\right]\right\}$