Answer
(-infinity, $0)$
Work Step by Step
$(2x-3)^2/x < 0$
Denominator is zero when $x=0$
$(2x-3)^2/x < 0$
$(2x-3)^2/x = 0$
$(2x-3)^2*x/x = 0*x$
$(2x-3)^2 = 0$
$\sqrt {(2x-3)^2} = \sqrt 0$
$2x-3 =0$
$2x=3$
$2x/2 = 3/2$
$x= 3/2$
(-infinity, $0)$
$(0, 3/2)$
$(3/2$, infinity)
Let $x=-3$, $x=1$, $x=2$
$x=-3$
$(2x-3)^2/x < 0$
$(2*-3-3)^2/-3 < 0$
$(-6-3)^2/-3 <0$
$(-9)^2/-3 < 0$
$81/-3 < 0$
$-27 < 0$ (true)
$x=1$
$(2x-3)^2/x < 0$
$(2*1-3)^2/1 < 0$
$(2-3)^2/1 < 0$
$(-1)^2/1 < 0$
$1/1 < 0$
$1 < 0$ (false)
$x=2$
$(2x-3)^2/x < 0$
$(2*2-3)^2/2 < 0$
$(4-3)^2/2 < 0$
$1^2/2 < 0$
$1/2 < 0$ (false)
