Using the following methods:
•Case method
•Test point method
•Conjunction/Disjunction method
x 1  0
2
Case Method
x 1x  1  0
Factor first
Now we should think in terms of multiplying.
We are multiplying two numbers and we get a
negative number.
Case I : First factor is positive and second factor is
negative
Case I : First factor is positive and second factor is
negative
x 1  0
x 1
AND
AND
x 1  0
x  1
This is impossible, as a number cannot be greater than
1 and smaller than negative 1. So we reject this case.
Case II : First factor is negative and second factor is
positive
x  1  0 AND x  1  0
x 1
AND
x  1
Case II is possible. Therefore, the graph is a
conjunction between -1 and 1 with shaded circles
over 1 and -1
Start from the factored form.
x 1x  1  0
Treat it as an equation and solve for x
x 1x  1  0
x  1 or x  1
Plot your zeros on a number line.
Place closed/shaded circles over our zeroes since the
problem was less than or equal to.
There are 3 regions we have to consider:
The region before -1
The region after 1
The region between -1 and 1
For each region, choose a test point.
For the region before -1, choose x = -2 (we could
have chosen any number less than -1)
Substitute your test point into the factored
expression:
 2 1 2  1
  3 1  3
The number we get is irrelevant, we only care about the sign.
So any point chosen before -1 will give a positive number.
We’ll mark this on our number line.
Choose a test point for the region between -1 and 1. 0 is a nice
choice here.
0 10  1
  11  1
This tells us than any number chosen in this region will yield a
negative number.
Let’s mark that on our number line.
For the last region of greater than 1, choose x = 2. Again, any
number greater than 1 would suffice.
2 12  1  13  3
This tells us than any number in this region will yield a positive
number. Like before, let’s mark this on our number line.
Now let’s go back to our factored form.
x 1x  1  0
The inequality sign tells us that our solution set is a number x
that will make the expression negative (less than or equal to 0)
If we look at our graph, the only region that happens is the
region between -1 and 1
Therefore, we choose this region. Our final graph then is:

This final method is the fastest, but use it very
cautiously
Start with the factored form, just like all other methods
x 1x  1  0
Treat it like an equation, like the test point method, and plot
your zeroes.
Now look back at your inequality in factored form and recall
absolute value inequalities. Greater than and greater than or
equal to meant your graph will be a disjunction. Less than or
less than or equal to implied a conjunction.
We have a conjunction, therefore we shade the region between
our “zeroes” (-1 and 1 here) as this would be a conjunction.
Our final graph then is a conjunction between -1 and 1
User Beware! This method can be faulty if the leading
coefficient of your quadratic inequality is negative!
Example – Solve and graph the following:
 x  5x  6  0
2
 x  2x  3  0
Treating like an equation and solving we get:
x  2 or x  3
Looking back at the original, this suggests that the graph is a
disjunction (because it is greater than 0). But the actual graph is
a conjunction between 2 and 3
To ensure that the last method will always work, we need to
make sure that our coefficient of x^2 is positive. We can do
this by multiplying by a negative one.
1 x
2

 5x  6  0 1
Remember: when dividing or multiplying by a negative
number, we flip the inequality sign
x  5x  6  0
2
x  2x  3  0
<
Treating like an equation, we get that x = 2 or x = 3
But this time, we see that our graph is really a conjunction
(since our expression is less than zero).