Transcript No Slide Title

Solving Quadratic Inequalities – Algebraic Method
• A quadratic inequality takes one of the following
forms:
ax  bx  c  0
2
ax  bx  c  0
2
ax  bx  c  0
2
ax  bx  c  0
2
• To solve the inequality means to find all values of the
variable for which the inequality is true.
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• Example 1
Use the algebraic method to solve the quadratic
inequality:
x  12  x
2
There are four steps to solve this quadratic inequality
by hand.
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x  12  x
2
Step 1: Replace the inequality sign with an equals sign
and solve the resulting equation.
x  12  x
2
x  x  12  0
2
 x  4  x  3  0
x   4, 3
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x   4, 3
These two values are called the boundary points.
Step 2: Place the boundary points on a number line.
4
3
Note that the boundary points separate the line into
three intervals.
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Step 3: Try a test number from each interval in the
original inequality.
3
4
True
5
False
0
True
4
x  12  x x  12  x x  12  x
2
 5 
2
2
2
 12  0 5  12   04   12   4 
2
2
25  12  5 0  12  0 16  12  4
25  17
0  12
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16  8
Step 4: Write the solution based on all True intervals.
4
True
5
3
False
0
True
4
   ,  4    3,  
We used a theorem here that states that if any value in
an interval is a solution in the original inequality, then
all values in that interval are solutions.
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   ,  4    3,  
Note that the original inequality was strictly
greater-than (no equal sign) …
x  12  x
2
This is why parentheses
were used next to the -4
and 3, rather than brackets.
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• Example 2
Use the algebraic method to solve the quadratic
inequality:
x  6 x  8
2
There are four steps to solve this quadratic inequality
by hand.
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x  6 x  8
2
Step 1: Replace the inequality sign with an equals sign
and solve the resulting equation.
x  6 x  8
2
x  6x  8  0
2
 x  4 x  2  0
x   4,  2
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x   4,  2
These two values are the boundary points.
Step 2: Place the boundary points on a number line.
2
4
Again, the boundary points separate the line into three
intervals.
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Step 3: Try a test number from each interval in the
original inequality.
2
4
False
5
True
3
False
0
x  6 x   8x  6 x  x8  6 x   8
2
 5 
2
2
2
 6  53 8 6   3 0  
 86  0    8
2
2
25  30   89  18   8 0  0   8
5  8
9  8
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0  8
Step 4: Write the solution based on all True intervals.
2
4
False
5
True
3
[  4,  2]
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False
0
  4,  2 
Note that the original inequality was less-than or
equal to.
x  6 x  8
2
Since the equal was
included, brackets are used
in the solution.
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