Answer
$(0,∞)$
Work Step by Step
$(x+1)^2/5x > 0$
The numerator is zero when $x=-1$, and the denominator is zero when $x=0$.
We have three sections: $(−∞,-1)$, $(-1,0)$, and $(0,∞)$. We need to test one value for x in each section to determine if the section would be a solution set. Since we have the $\gt$ sign, we include the end points and use brackets instead of parentheses.
Let $x=-2$, $x=-.5$, $x=1$
$x=-2$
$(x+1)^2/5x > 0$
$(-2+1)^2/5*-2 > 0$
$(-1)^2/-10 > 0$
$1/-10 >0$
$-1/10 > 0$ (false)
$x=-.5$
$(x+1)^2/5x > 0$
$(-.5+1)^2/5*-.5 > 0$
$(.5)^2 /-2.5 >0$
$.25/-2.5 > 0$
$-1/10 > 0$ (false)
$x=1$
$(x+1)^2/5x >0$
$(1+1)^2/5*1 >0$
$2^2/5 >0$
$4/5 > 0$ (true)
