11.2 Solving
Quadratic Equations
by Completing the
Square
Objective 1
Solve quadratic equations by
completing the square when the
coefficient of the second-degree term
is 1.
Slide 11.2-3
Solve quadratic equations by completing the square when
the coefficient of the second-degree term is 1.
The methods studied so far are not enough to solve the equation
x2 + 6x + 7 = 0.
If we could write the equation in the form (x + 3)2 equals a constant, we
could solve it with the square root property discussed in Section 11.1. To do
that, we need to have a perfect square trinomial on one side of the
equation.
Recall from Section 5.4 that th perfect square trinomial has the form
x2 + 6x + 9 can be factored as (x + 3)2.
Slide 11.2-4
Solve quadratic equations by completing the square when
the coefficient of the second-degree term is 1. (cont’d)
The process of changing the form of the equation from
x2 + 6x + 7 = 0
to
(x + 3)2 = 2
is called completing the square. Completing the square changes only the
form of the equation.
Completing the square not only provides a method for solving quadratic equations,
but also is used in other ways in algebra (finding the coordinates of the center of a
circle, finding the vertex of a parabola, and so on).
Slide 11.2-5
CLASSROOM
EXAMPLE 1
Rewriting an Equation to Use the Square Root Property
Solve x2 – 20x + 34 = 0.
Solution:
x 2  20 x  34
x 2  20 x  _____
x 2  20 x  100  34  100
2kx  20x
 x  10 
2
 66
x  10  66 or x  10   66
x  10  66 or
10 
x  10  66
66
2k  20
k  10
k 2  100

Slide 11.2-6
CLASSROOM
EXAMPLE 2
Completing the Square to Solve a Quadratic Equation
Solve x2 + 4x = 1.
Solution:
x2  4 x  1
x2  4 x  4  1  4
 x  2
2
5
x2 5
x  2 
5
or
x  2 
5
2  5
Slide 11.2-7
Solve quadratic equations by completing the square when
the coefficient of the second-degree term is 1.
Completing the Square
To solve ax2 + bx + c = 0 (a ≠ 0) by completing the square, use
these steps.
Step 1 Be sure the second-degree (squared) term has
coefficient 1. If the coefficient of the squared term is one,
proceed to Step 2. If the coefficient of the squared term is
not 1 but some other nonzero number a, divide each side of
the equation by a.
Step 2 Write the equation in correct form so that terms with
variables are on one side of the equals symbol and the
constant is on the other side.
Step 3 Square half the coefficient of the first-degree (linear)
term.
Slide 11.2- 7
Solve quadratic equations by completing the square when
the coefficient of the second-degree term is 1.
Completing the Square (continued)
Step 4 Add the square to each side.
Step 5 Factor the perfect square trinomial. One side should now
be a perfect square trinomial. Factor it as the square of a
binomial. Simplify the other side.
Step 6 Solve the equation. Apply the square root property to
complete the solution.
Steps 1 and 2 can be done in either order. With some equations, it is more convenient
to do Step 2 first.
Slide 11.2- 8
CLASSROOM
EXAMPLE 3
Solving a Quadratic Equation by Completing the Square (a = 1)
Solve x2 + 3x – 1 = 0.
Solution:
x2 + 3x = 1
Completing the square.
2
3 9
  
4
2
Add the square
to each side.
x2
9
9
+ 3x + = 1 +
4
4
2
3  13

x  
2
4

Slide 11.2- 9
CLASSROOM
EXAMPLE 3
Solving a Quadratic Equation by Completing the Square (a = 1) (cont’d)
Use the square root property.
3
13
x 
2
4
3
13
x 
2
2
3  13
x
2
Check that the solution set is
or
or
or
3
13
x  
2
4
3
13
x 
2
2
3  13
x
2
 3  13 

.
2


Slide 11.2- 10
Objective 2
Solve quadratic equations by
completing the square when the
coefficient of the second-degree term
is not 1.
Slide 11.2-12
Solve quadratic equations by completing the square when the
coefficient of the second-degree term is not 1.
Solving a Quadratic Equation by Completing the Square
Step 1: Be sure the second-degree term has a coefficient of 1. If
the coefficient of the second-degree term is 1, go to Step 2. If
it is not 1, but some other nonzero number a, divide each
side
of the equation by a.
Step 2: Write in correct form. Make sure that all variable terms are on the
one side of the equation and that all constant terms are on the
other.
Step 3: Complete the square. Take half of the coefficient of the first degree
term, and square it. Add the square to both sides of the equation.
Factor the variable side and combine like terms on the other side.
Step 4: Solve the equation by using the square root property.
If the solutions to the completing the square method are rational numbers, the
equations can also be solved by factoring. However, the method of completing
the square is a more powerful method than factoring because it allows us to solve
any quadratic equation.
Slide 11.2-13
CLASSROOM
EXAMPLE 4
Solving a Quadratic Equation by Completing the Square
Solve 4x2 + 8x -21 = 0.
Solution:
4 x 2  8 x  21  0
21
2
x  2x   0
4
21
2
x  2x 
4
21 4
2
x  2x 1  
4 4
25
2
x  2x 1 
4
 x  1 
2
25
x 1 
4
5
x 1 
2
x
3
2
25
4
25
or x  1  
4
5
x

1


or
2
or
x
7
2
 7 3
 , 
 2 2
Slide 11.2-14
CLASSROOM
EXAMPLE 5
Solving a Quadratic Equation by Completing the Square (a ≠ 1)
Solve 3x2 + 6x – 2 = 0.
Solution:
3x2 + 6x = 2
2
2
x  2x 
3
Completing the square.

1
2
 2
2
 1  1
2
2
x  2x 1  1
3
2
5
 x  1 
3
2
Slide 11.2- 14
CLASSROOM
EXAMPLE 5
Solving a Quadratic Equation by Completing the Square (a ≠ 1) (cont’d)
Use the square root property.
5
x 1 
3
5
x  1 
3
15
x  1 
3
3  15
x
3
Check that the solution set is
or
3
5
x 1  
2
3
or
5
x  1 
3
15
or
x  1 
3
3  15
or
x
3
 3  15 

.
3


Slide 11.2- 15
CLASSROOM
EXAMPLE 6
Solve for Nonreal Complex Solutions
Solve the equation.
5 x 2  15 x  12  0
Solution:
5 x 2  15 x  12
12
2
x  3x  
5
Complete the square.
2
9
 3
  3      
4
 2
1
2
2
9
12 9
x  3x    
4
5 4
2
3
3

x  
2
20

2
Slide 11.2- 16
CLASSROOM
EXAMPLE 6
Solve for Nonreal Complex Solutions (cont’d)
3
3
x  
2
20
3 i 3 5
x 

2
20 5
3 i 15
x 
2
10
3
3
or x    
2
20
3 i 3 5
or x  

2
20
5
3 i 15
or x  
2
10
3 i 15
3 i 15
x 
or x  
2 10
2 10
 3
15 
The solution set is  
i.
 2 10 
Slide 11.2- 17
Objective 3
Simplify the terms of an equation
before solving.
Slide 11.2-19
CLASSROOM
EXAMPLE 7
Simplifying the Terms of an Equation before Solving
Solve (x + 6)(x + 2) = 1.
Solution:
x 2  8 x  12  1
( x  4)  5 or ( x  4)   5
x  4  5 or x  4  5
x 2  8 x  16  11
x  8x  16  5
 x  4
2
5
4  5
Slide 11.2-20