Inequalities
Objectives
• Understand the laws for working with
inequality symbols
• Solve linear inequalities
• Solve quadratic inequalities
Linear inequalities
3 x  10  10 x  11
3 x  10 x  11  10
 7 x  21
x3
• You can add or subtract a
number on both sides of an
inequality
• You can multiply or divide an
inequality by a positive
number
• You can multiply or divide an
inequality by a negative
number, but you must change
the direction of the inequality
Quadratic Inequalities
ax 2  bx  c  a( x  p)( x  q)  a( x  r ) 2  s
usual form
factor form
completed square form
Easier form to use
with inequalities
An interval (AND) example: Method 1
y is less than 0
Coefficients of x2 is
positive  the parabola
bents upwards
y  x  2x  4  0
Intersects the x axis at 2 and 4
Please note
that 2 is less
than 4
Be careful, this does not
make sense; x can’t be
both (AND) greater than
7 and less than 3. Ok for
OR condition
7< x <3
Please note
that 7 is not
less than 3
X must be
both greater
than 2 AND
less than 4
2<x<4
interval
Method 2 (tabular method)
y  x  2x  4  0
x<2
x=2
2<x<4
x=4
x>4
x-2
-
0
+
+
+
x-4
-
-
-
0
+
(x-2)(x-4)
+
0
-
0
+
x  2x  4  0
X > 2 AND X < 4
For what values of input
the output will be less
than 0?
x
f x 
3x 2  3x  18  3x  6x  3
x<-2
x=-2
-2<x<3
x+2
-
0
+
x-3
-
(x+2)(x-3)
+
0
x=3
y
The graph is below x axis
for input values between -2
and 3. in other words the
output of the function is
negative for input values
between -2 and 3
x>3
+
-
0
+
-
0
+
Range (interval) bound
The graph shows the output (y) of
the function for various inputs (x)
x  2x  3  0
X > -2 AND X < 3
Range (interval) bound
OR instead of AND
Either condition would satisfy
For what values of input
the output will be greater
than 0?
x
f x 
3x 2  3x  18  3x  6x  3
X < -2
X>3
The graph is above x axis if
input values are < -2 or > 3.
in other words the output of
the function is positive for
input values < -2 and > 3
y
Not range (interval) bound
x<-2
x=-2
-2<x<3
x+2
-
0
+
x-3
-
(x+2)(x-3)
+
0
x=3
x>3
+
-
0
+
-
0
+
x  2x  3  0
X < -2 OR X > 3
Not range (interval) bound
The graph shows the output (y) of
the function for various inputs (x)
If two products are positive, then either they both
should be positive or they both should be negative
( x  3)(3x  6)  0
Scenario 1
Scenario 2
If two products are negative, then one should be
negative and the other should be positive
( x  3)(3x  6)  0
Scenario 1
Scenario 2
(3x  6)  0  x  2
(3x  6)  0  x  2
(3x  6)  0  x  2
(3x  6)  0  x  2
( x  3)  0  x  3
( x  3)  0  x  3
( x  3)  0  x  3
( x  3)  0  x  3
This means x > 3
This means x < -2
This is a OR condition (not range/interval bound)
This means -2< x > 3
This means x < -2
x must be both greater than 3
and less than -2. This AND
condition is impossible.
Another Example: OR criteria (not range/interval bound)
Coefficients of x2 is
negative  the parabola
bents downwards (vertex
at the top)
Y>0
Y<0
( x  1)(5  x)  0
Intersects the x axis at -1 and 5
x  1
Critical values
Either condition would satisfy
OR instead of AND
x5
x2  a2
a0
x 2  a 2  ( x  a)( x  a)  0
It is easier to assign
simple values to a
Critical values are x= –a and x =+a
Ex: x = -4
Ex: x = 2
Ex: x = 4
x<-3
x=-3
-3<x<3
x=3
x>3
x-3
-
-
-
0
+
x+3
-
0
+
+
+
(x-3)(x+3)
+
0
-
0
+
a  x  a
x2  a2
Both are equivalent
Another Approach
If two products are negative, one of them MUST be
negative and the other MUST be positive
(2 x  1)( x  3)  0
Scenario 1
Scenario 2
(2 x  1)  0  x   1 2
(2 x  1)  0  x   1 2
( x  3)  0  x  3
( x  3)  0  x  3
x must be both greater
than 3 and less than -1/2.
This AND condition is
impossible.
x must be both less than 3
and greater than -1/2.
If two products are positive, then
either they both should be positive or
they both should be negative
(2 x  1)( x  3)  0
Scenario 1
(2 x  1)  0  x   1
( x  3)  0  x  3
This means x > 3
2
Scenario 2
(2 x  1)  0  x   1
2
( x  3)  0  x  3
This means x < -1/2
This is a OR condition (not range/interval bound)
Don’t have a calculator and can’t factorize the equation or your mean teacher
asked you to solve algebraically: complete the square
Square means a positive value
2 x 2  8x  11  0
 2( x  2) 2  3  0
There is no value of x
for which this condition
is possible
Equals 0 when x = 2 ;
the smallest value
2 x 2  8 x  5  2( x  2) 2  3  0  x  2 
2

We saw this a
couple a slides ago.
3
3
 x2
2
2
 2
3
2
3
3
 x  2
2
2
x 2  a 2  a  x  a
We have two functions: y1, y2. Find the
range of x values (domain) so that y1 > y2
y2  x  4
3x
y1  1 
2
1
3x
 x4 x  2
2
y2  y1
Graphical proof: the graphs
seem to intersect at (2, -2).
When x is smaller than 2, y1
is greater than y2
x  5  2   x  5  4
2
Scenario 1
x 2  10 x  21  0
x  7   0
 x  3  0
x 2  7 x  3 x  21  0
Scenario 2
x 2  10 x  25  4  0
x x  7   3 x  7   0
x  7 x  3  0
x  7   0
 x  3  0
3 x 7
x5  2
  x  5  2
x  5  2 or   x  5  2
This is impossible
3 x 7
x7
x  5  2
x3