Chapter 1.4
Quadratic Equations
Quadratic Equation in One Variable
An equation that can be written in the form
ax2 + bx + c = 0
where a, b, and c, are real numbers,
is a quadratic equation
A quadratic equation is a second-degree
equation—that is, an equation with a
squared term and no terms of greater
degree.
x2 =25,
4x2 + 4x – 5 = 0,
3x2 = 4x - 8
A quadratic equation written in the form
ax2 + bx + c = 0 is in standard form.
Solving a Quadratic Equation
Factoring is the simplest method of solving a
quadratic equation (but one not always easily
applied).
This method
depends on the
zero-factor
property.
Zero-Factor Property
If two numbers have a product of 0 then at
least one of the numbers must be zero
If ab= 0
then a = 0 or b = 0
Example 1. Using the zero factor property.
Solve 6x2 + 7x = 3
A quadratic equation of the form x2 = k can
also be solved by factoring.
x2 = k
x2 – k=0
x  k x  k   0
x k 0
x k
or x  k  0
or x   k
T hisprovesthesquare root property.
Square root property
If x2 = k, then
x k
or x   k
Example 2 Using the Square Root Property
Solve each quadratic equation.
x2 = 17
Example 2 Using the Square Root Property
Solve each quadratic equation.
x2 = -25
Example 2 Using the Square Root Property
Solve each quadratic equation.
(x-4)2 = 12
Completing the Square
Any quadratic equation can be solved by the
method of completing the square.
Example 3 Using the Method of Completing
the Square, a = 1
Solve x2 – 4x – 14 = 0
Example 3 Using the Method of Completing
the Square, a = 1
Solve x2 – 4x – 14 = 0
x 2  4x  14
Example 3 Using the Method of Completing
the Square, a = 1
4
Solve x2 – 4x – 14 = 0
x  4x  14
2
2
Example 3 Using the Method of Completing
the Square, a = 1
4
Solve x2 – 4x – 14 = 0
 2
x  4x  14
2
2
Example 3 Using the Method of Completing
the Square, a = 1
4
Solve x2 – 4x – 14 = 0
 2
x  4x  14
2
2
 2
2
Example 3 Using the Method of Completing
the Square, a = 1
4
Solve x2 – 4x – 14 = 0
 2
x  4x  14
2
2
 2
2
 4
Example 3 Using the Method of Completing
the Square, a = 1
4
Solve x2 – 4x – 14 = 0
 2
x  4x  14
2
x  4x  4  14  4
 2
2
2
2
 4
Example 3 Using the Method of Completing
the Square, a = 1
4
Solve x2 – 4x – 14 = 0
 2
x  4x  14
2
x  4x  4  14  4  18
2
2
 2
2
 4
Example 3 Using the Method of Completing
the Square, a = 1
4
Solve x2 – 4x – 14 = 0
 2
x  4x  14
2
x  4x  4  14  4  18
2
( x  2)
2
2
 2
2
 4
Example 3 Using the Method of Completing
the Square, a = 1
4
Solve x2 – 4x – 14 = 0
 2
x  4x  14
2
x  4x  4  14  4  18
2
( x  2)  18
2
2
 2
2
 4
Example 3 Using the Method of Completing
the Square, a = 1
4
Solve x2 – 4x – 14 = 0
 2
x  4x  14
2
x  4x  4  14  4  18
2
( x  2)  18
2
x  2   18
2
 2
2
 4
Example 3 Using the Method of Completing
the Square, a = 1
4
Solve x2 – 4x – 14 = 0
 2
x  4x  14
2
x  4x  4  14  4  18
2
( x  2)  18
2
x  2   18   9 2
2
 2
2
 4
Example 3 Using the Method of Completing
the Square, a = 1
4
Solve x2 – 4x – 14 = 0
 2
x  4x  14
2
x  4x  4  14  4  18
2
2
 2
( x  2)  18
2
x  2   18   9 2  3 2
2
 4
Example 3 Using the Method of Completing
the Square, a = 1
4
Solve x2 – 4x – 14 = 0
 2
x  4x  14
2
x  4x  4  14  4  18
2
2
 2
( x  2)  18
2
x  2   18   9 2  3 2
x  2  3 2
2
 4
Example 3 Using the Method of Completing
the Square, a = 1
4
Solve x2 – 4x – 14 = 0
 2
x  4x  14
2
x  4x  4  14  4  18
2
2
 2
( x  2)  18
2
x  2   18   9 2  3 2
x  2  3 2
x  23 2
2
 4
Example 4 Using the Method of Completing
the Square, a ≠1
Solve 9x2 – 12x – 1 = 0
Example 4 Using the Method of Completing
the Square, a ≠1
Solve 9x2 – 12x – 1 = 0
12
1
x  x 0
9
9
2
Example 4 Using the Method of Completing
the Square, a ≠1
Solve 9x2 – 12x – 1 = 0
12
1
x  x 0
9
9
4
1
2
x  x 0
3
9
2
Example 4 Using the Method of Completing
the Square, a ≠1
Solve 9x2 – 12x – 1 = 0
12
1
x  x 0
9
9
4
1
2
x  x

3
9
2
Example 4 Using the Method of Completing
the Square, a ≠1
Solve 9x2 – 12x – 1 = 0
12
1
x  x 0
9
9
4
1
2
x  x

3
9
2
 1  4 
 2   3 

 
2
Example 4 Using the Method of Completing
the Square, a ≠1
Solve 9x2 – 12x – 1 = 0
12
1
2
x  x 0
9
9
4
1
2
x  x

3
9
 1  4 
 2   3 

 
 2
  
 3
2
2
Example 4 Using the Method of Completing
the Square, a ≠1
Solve 9x2 – 12x – 1 = 0
12
1
2
x  x 0
9
9
4
1
2
x  x

3
9
 1  4 
 2   3 

 
 2
  
 3
4

9
2
2
Example 4 Using the Method of Completing
the Square, a ≠1
Solve 9x2 – 12x – 1 = 0
12
1
2
x  x 0
9
9
4
1
2
x  x

3
9
4
4 1 4
2
x  x  
3
9 9 9
 1  4 
 2   3 

 
 2
  
 3
4

9
2
2
Example 4 Using the Method of Completing
the Square, a ≠1
Solve 9x2 – 12x – 1 = 0
12
1
2
x  x 0
9
9
4
1
2
x  x

3
9
4
4 1 4 5
2
x  x   
3
9 9 9 9
 1  4 
 2   3 

 
 2
  
 3
4

9
2
2
Example 4 Using the Method of Completing
the Square, a ≠1
4
4 5
x  x 
3
9 9
2
Example 4 Using the Method of Completing
the Square, a ≠1
4
4 5
x  x 
3
9 9
2
2
5
 2
x -  
9
 3
Example 4 Using the Method of Completing
the Square, a ≠1
4
4 5
x  x 
3
9 9
2
2
5
 2
x -  
9
 3
2
5
x- 
3
9
Example 4 Using the Method of Completing
the Square, a ≠1
4
4 5
x  x 
3
9 9
2
2
5
 2
x -  
9
 3
2
5
x- 
3
9
2
5
x 
3
9
Example 4 Using the Method of Completing
the Square, a ≠1
4
4 5
x  x 
3
9 9
2
2
5
 2
x -  
9
 3
2
5
x- 
3
9
2
5
x 
3
9
2
5
x 
3
9
Example 4 Using the Method of Completing
the Square, a ≠1
4
4 5
x  x 
3
9 9
2
2
5
 2
x -  
9
 3
2
5
x- 
3
9
2
5
x 
3
9
2
5
x 
3
9
2
5
x 
3
3
Example 4 Using the Method of Completing
the Square, a ≠1
4
4 5
x  x 
3
9 9
2
2
5
 2
x -  
9
 3
2
5
x- 
3
9
2
5
x 
3
9
2
5
x 
3
9
2
5
x 
3
3
2 5
x
3
The Quadratic Formula
ax  bx  c  0
2
Watch the derivation
x
b  b  4 ac
2a
2
Example 5 Using the Quadratic Formula
(Real Solutions)
Solve x2 -4x = -2
Example 6 Using the Quadratic Formula
(Non-real Complex Solutions)
Solve 2x2 = x – 4
Example 7 Solving a Cubic Equation
Solve x3 + 8 = 0
Example 8 Solving a Variable That is Squared
Solve for the specified variable.
A
d
4
2
, for d
Example 8 Solving a Variable That is Squared
Solve for the specified variable.
rt  st  k (r  0), for t
2
The Discriminant The quantity under the
radical in the quadratic formula,
b2 -4ac, is called the discriminant.
 b  b  4ac
x
2a
2
Discriminant
Then the numbers a, b, and c are integers, the value
of the discriminant can be used to determine
whether the solution of a quadratic equation are
rational, irrational, or nonreal complex numbers, as
shown in the following table.
Discriminant Number of Solutions Kind of Solutions
Positive
Two
Rational
Two
Irrational
(Perfect Square)
Positive
(but not a Perfect Square)
Zero
One
Rational
(a double solution)
Negative
Two
Nonreal complex
Example 9 Using the Discriminant
Determine the number of solutions and tell
whether they are rational, irrational, or nonreal
complex numbers.
2
2
 b  b  4ac
5x + 2x – 4 = 0
x
2a
a
b
c
 ( )  ( )  4( )( )
x
2( )
2
Example 9 Using the Discriminant
Determine the number of solutions and tell
whether they are rational, irrational, or nonreal
complex numbers.
2
2
 b  b  4ac
x – 10x = -25
x
2a
a
b
c
 ( )  ( )  4( )( )
x
2( )
2
Example 9 Using the Discriminant
Determine the number of solutions and tell
whether they are rational, irrational, or nonreal
complex numbers.
2
2
 b  b  4ac
2x – x + 1 = 0
x
2a
a
b
c
 ( )  ( )  4( )( )
x
2( )
2
Homework 1.4 # 1-79