Answer
$(−∞,0)$
Work Step by Step
$(2x-3)^2/x < 0$
$2x-3=0$
$2x-3+3=0+3$
$2x=3$
$2x/2 =3/2$
$x=3/2$
The numerator is zero when $x=3/2$, and the denominator is zero when $x=0$.
We have three sections: $(−∞,0)$, $(0,3/2)$, and $(3/2,∞)$. 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 $\lt$ sign, we exclude the end points and use parentheses instead of brackets.
Let $x=-1$, $x=1$, $x=3$
$x=-1$
$(2x-3)^2/x < 0$
$(2*-1-3)^2/-1 < 0$
$(-2-3)^2/-1<0$
$(-5)^2/-1<0$
$25/-1 <0$
$-25 <0$ (true)
$x=1$
$(2x-3)^2/x < 0$
$(2*1-3)^2/1 < 0$
$(2-3)^2 <0$
$(-1)^2 <0$
$1 <0$ (false)
$x=3$
$(2x-3)^2/x < 0$
$(2*3-3)^2/3 < 0$
$(6-3)^2/3<0$
$3^2/3 <0$
$9/3 <0$
$3<0$ (false)
