Start by setting the inequality to zero and by solving for .

Now, plot these two points on to a number line.

Notice that these two numbers effectively divide up the number line into three regions:

, , and 

Now, choose a number in each of these regions and put it back in the factored inequality to see which cases are true.

For , let 

Since this  is not less than , the solution to this inequality cannot lie in this region.

For , let .

Since this will make the inequality true, the solution can lie in this region.

Finally, for , let 

Since this number is not less than zero, the solution cannot lie in this region.

Thus, the solution to this inequality is 

First, set the inequality to zero and solve for .

Now, plot these two numbers on to a number line.

Notice how these numbers divide the number line into three regions:

Now, you will choose a number from each of these regions to test to plug back into the inequality to see if the inequality holds true.

For , let 

Since this is not less than zero, the solution to the inequality cannot be found in this region.

For , let 

Since this is less than zero, the solution is found in this region.

For , let 

Since this is not less than zero, the solution is not found in this region.

Then, the solution for this inequality is 

Start by changing the less than sign to an equal sign and solve for .

Now, plot these two numbers on a number line.

Notice how the number line is divided into three regions:

Now, choose a number fromeach of these regions to plug back into the inequality to test if the inequality holds.

For , let 

Since this number is not less than zero, the solution cannot be found in this region.

For , let 

Since this number is less than zero, the solution can be found in this region.

For  let .

Since this number is not less than zero, the solution cannot be found in this region.

Because the solution is only negative in the interval , that must be the solution.

First, set the inequality to zero and solve for .

Now, plot these two numbers on to a number line.

Notice how these numbers divide the number line into three regions:

Now, you will choose a number from each of these regions to test to plug back into the inequality to see if the inequality holds true.

For , let 

Since this solution is greater than or equal to , the solution can be found in this region.

For , let 

Since this is less than or equal to , the solution cannot be found in this region.

For , let 

Since this is greater than or equal to , the solution can be found in this region.

Because the solution can be found in every single region, the answer to this inequality is 

First, we can factor the quadratic to give us a better understanding of its graph. Factoring gives us: . Now we know that the quadratic has zeros at  and . Furthermore this information reveals that the quadratic is positive. Using this information, we can sketch a graph like this: 

We can see that the parabola is below the x-axis (in other words, less than ) between these two zeros  and .

The only x-value satisfying the inequality  is .

The value of  would work if the inequality were inclusive, but since it is strictly less than instead of less than or equal to , that value will not work.