Question

Solve each inequality algebraically.

$\frac{(x-1)(x+1)}{x}\le 0$

Short answer

Solution to the inequality $\frac{(x-1)(x+1)}{x}\le 0$is $(-\infty ,-1]\cup (0,1]$.

Step-by-step solution

Step 1. Given information 

We have been given an inequality $\frac{(x-1)(x+1)}{x}\le 0$.

We have to solve this inequality algebraically.

Step 2. Determine the real numbers at which the expression f equals zero and at which the expression f is undefined.  

Assume $f\left(x\right)=\frac{(x-1)(x+1)}{x}$.

So, $$\begin{gathered} (x-1)(x+1)=0 \\ x=-1,1 \end{gathered}$$

Also,$x=0$

Step 3. Form the intervals  

Using the values of x found in previous step, we can divide the real numbers in the intervals:

$(-\infty ,-1)\cup (-1,0)\cup (0,1)\cup (1,\infty )$

Step 4. Select a number in each interval and evaluate f at the number  

Create the following table:

Interval	$(-\infty ,-1)$	$(-1,0)$	$(0,1)$	$(1,\infty )$
Number chosen	$-3$	$-0.5$	$0.5$	2
Value of f	$f(-3)=-\frac{8}{3}$	$f(-0.5)=\frac{3}{2}$	$f(0.5)=-\frac{3}{2}$	$f\left(2\right)=\frac{3}{2}$
Conclusion	negative	positive	negative	positive

Step 5. Identify the interval 

Since we want to know where f is negative or zero, we conclude that $f\left(x\right)\le 0$ in the interval .