Here we have a quotient inequality of a special nature. We can rearrange it and multiply through by the denominator since the modulus of a function is always positive.

Recall one law of indices .

 or

Solution set

Applying the difference of two squares we have:

Multiplying through by -1 we have

Solution set

Applying the difference of two squares we have:

Multiplying through by -1 we have

Solution set

We may solve this problem by proceeding thus

Recall that

• If
•  or

a) If 

b) If
 or
 or
 or
c) If

Rearranging and factorizing we have

Multiplying the numerator and denominator by , the denominator;
Hence
The zeros are , or

The solution set is

We rearrange and simplify thus

Multiplying the numerator and the denominator by the denominator;

Since the denominator is always positive then the numerator in this problem must also be positive.
Hence
The zeros are  or

The solution set is

Rearranging and factorizing we have

Multiplying the numerator and denominator by , the denominator;
Hence
The zeros are , or

The solution set is

n

We rearrange and simplify as we did in problem 6 above.

 or
Dividing through by -2 , the inequality sign is reversed.

Since the denominator is always positive for all real values of , it implies the numerator must be negative or less than zero

The zeros are  or

The solution set is

Factorizing we have
The zeros are  or