Answer
SOLUTION
$\frac{(2+x)(2-x)}{x}\lt0$
INTERVAL
(-2, 0) (2, $\infty)$
GRAPH
In the real number line there should be the numbers -2, 0 and 2. In those three there should be an opne circle. The you connect -2 and 0. The you draw an arrow from 2 to $\infty$.
Work Step by Step
$\frac{4}{x}\lt(x)$
$\frac{4}{x}-x\lt0$
$\frac{4}{x}-\frac{x^{2}}{x}\lt0$
$\frac{4-x^{2}}{x}\lt0$
$\frac{(2+x)(2-x)}{x}\lt0$
In this inequality the answer can’t be equal to 0 therefore the answer go in parenthses.
The solutions of the numerator are
x=-2 and x=2
The solutions of the denominator is
x=0
INTERVAL
$(-\infty, -2) (-2,0) (0,2) (2,\infty)$
Test the intervals to know which one is the solution
$(-\infty, -2)$ $y(-4)=\frac{(2-4)(2+4)}{-4}\lt0$ $y(-4)=3$
The result is greater than 0 so it is not a solution.
(-2,0) $y(-1)=\frac{(2-1)(2+1)}{-1}\lt0$ $y(-1)=-3$
The result is less than 0 so the interval is a solution.
(0,2) $ y(1)=\frac{(2+1)(2-1)}{1}\lt0$ $y(1)=3$
It is not a solution
$(2, \infty)$ $y(4)=\frac{(2+4)(2-4)}{4}\lt0$ $y(4)= -3$
It is a solution.
SOLUTION (-2,0) $(2,\infty)$
GRAPH
In the real number line there should be the numbers -2, 0 and 2. In those three there should be an opne circle. The you connect -2 and 0. The you draw an arrow from 2 to $\infty$.
