This document provides instructions for solving quadratic equations using the method of completing the square. It discusses how to complete the square to create a perfect square trinomial, set it equal to a constant, and take the square root. Several examples are worked through step-by-step to demonstrate how to isolate terms, calculate the quantity to be added inside the square, and solve for the variable. The quadratic formula is also introduced as an alternative method using the standard form of the quadratic equation.

1. Martin-Gay, Developmental Mathematics 1
Solving Quadratic Equations by Completing
the Square

2. Martin-Gay, Developmental Mathematics 2
Completing the Square
This method will work to solve ALL quadratic
equations; however,
it is “messy” to solve quadratic equations by
completing the square if a ≠ 1 and/or b is an odd
number.
Completing the square is a great choice for solving
quadratic equations if a = 1 and b is an even
number.
It involves creating a trinomial that is a perfect
square, setting the factored trinomial equal to a
constant, then using the square root property from
the previous lesson.

3. Martin-Gay, Developmental Mathematics 3
Solving a Quadratic Equation by Completing a Square
1) If the coefficient of x2 is NOT 1, divide both sides of
the equation by the coefficient.
2) Isolate all variable terms on one side of the equation.
3) Complete the square (half the coefficient of the x
term squared, added to both sides of the equation).
4) Factor the resulting trinomial.
5) Use the square root property. (extracting root)
6) Solve for the value of the variable.
Completing the Square

4. Martin-Gay, Developmental Mathematics 4
Solve by completing the square.
y2 + 6y = 8
y2 + 6y + 9 = 8 + 9
(y + 3)2 = 1
y = 3 ± 1
y = 4 or 2
y + 3 = ± = ± 1
1
Solving Equations
Example

5. Martin-Gay, Developmental Mathematics 5
Solve by completing the square.
y2 + y – 7 = 0
y2 + y = 7
y2 + y + ¼ = 7 + ¼
2
29
4
29
2
1





y
2
29
1
2
29
2
1 





y
(y + ½)2 = 4
29
Solving Equations
Example

6. Martin-Gay, Developmental Mathematics 6
Solve by completing the square.
2x2 + 14x – 1 = 0
2x2 + 14x = 1
x2 + 7x = ½
2
51
4
51
2
7





x
2
51
7
2
51
2
7 





x
x2 + 7x + = ½ +
4
49
4
49
(x + )2 =
4
51
2
7
Solving Equations
Example

7. Martin-Gay, Developmental Mathematics 7
 Example
a ≠ 1, b is not even
3x
2
– 5x + 2 = 0
2 5 2 0
3 3
x x
  
2 5 25 2 25
3 36 3 36
x x
   
2
5 1
6 36
x
 
 
 
 
 
5 1
6 6
x 
5 1
6 6
x 
5 1
6 6
x 
x = 1 OR x = ⅔

8. Martin-Gay, Developmental Mathematics 8
Solving Quadratic Equations by the
Quadratic Formula

9. Martin-Gay, Developmental Mathematics 9
The Quadratic Formula
Another technique for solving quadratic equations is to use
the quadratic formula.
The formula is derived from completing the square of a
general quadratic equation.

10. Martin-Gay, Developmental Mathematics 10
The Quadratic Formula
A quadratic equation written in standard form, ax2 + bx + c = 0, has
the solutions.
a
ac
b
b
x
2
4
2





11. Martin-Gay, Developmental Mathematics 11
The Quadratic Formula
Solve 11n2 – 9n = 1 by the quadratic formula.
11n2 – 9n – 1 = 0, so
a = 11, b = -9, c = -1






)
11
(
2
)
1
)(
11
(
4
)
9
(
9 2
n 


22
44
81
9


22
125
9
22
5
5
9 
Example

12. Martin-Gay, Developmental Mathematics 12






)
1
(
2
)
20
)(
1
(
4
)
8
(
8 2
x 



2
80
64
8



2
144
8



2
12
8 20 4
or , 10 or 2
2 2


x2 + 8x – 20 = 0 (multiply both sides by 8)
a = 1, b = 8, c = 20
8
1
2
5
Solve x2 + x – = 0 by the quadratic formula.
The Quadratic Formula
Example

13. Martin-Gay, Developmental Mathematics 13
The Quadratic Formula
Solve x(x + 6) = 30 by the quadratic formula.
x2 + 6x + 30 = 0
a = 1, b = 6, c = 30





)
1
(
2
)
30
)(
1
(
4
)
6
(
6 2
x 



2
120
36
6
2
84
6 


So there is no real solution.
Example