Completing the Square
• Objective: To complete a square for a
quadratic equation and solve by completing
the square
Steps to complete the square
• 1.) You will get an expression that looks
like this: AX²+ BX
• 2.) Our goal is to make a square such that
we have (a + b)² = a² +2ab + b²
• 3.) We take ½ of the X coefficient (Divide
the number in front of the X by 2)
• 4.) Then square that number
• Take half of the coefficient of ‘x’
• Square it and add it
To Complete the Square
x2 + 6x
3
9
x2 + 6x + 9 = (x + 3)2
Complete the square, and show what the
perfect square is:
x
x 12
2
 36
12
2

 x
x  2
6

x
y
y 14
2
 49
14
2

 y
y  2
7

y
y
y 10
2
 25
10
2

 y
y  2
5

y
x
x 5
2

4
25
5
2

 x
x
2
2
5







x
To solve by completing the square
• If a quadratic equation does not factor we
can solve it by two different methods
• 1.) Completing the Square (today’s lesson)
• 2.) Quadratic Formula (Monday’s lesson)
Steps to solve by completing the square
1.) If the quadratic does not factor, move the constant to the
other side of the equation
Ex: x²-4x -7 =0 x²-4x=7
2.) Work with the x²+ x side of the equation and complete
the square by taking ½ of the coefficient of x and squaring
Ex. x² -4x 4/2= 2²=4
3.) Add the number you got to complete the square to both
sides of the equation
Ex: x² -4x +4 = 7 +4
4.)Simplify your trinomial square
Ex: (x-2)² =11
5.)Take the square root of both sides of the equation
Ex: x-2 =±√11
6.) solve for x
Ex: x=2±√11
Ex 1: Solve by Completing the Square
2
6 16 0
x x
  
2
6 16
x x
 
+9 +9
2
6 9 25
x x
  
 
2
3 25
x  
 
3 5
x   
3 5
x   8
x  2
x  
Ex 2: Solve by Completing the Square
2
2 5 0
x x
  
2
2 5
x x
 
+1 +1
2
2 1 6
x x
  
 
2
1 6
x  
 
1 6
x   
1 6
x  
Ex 3: Solve by Completing the Square
2
10 4 0
x x
  
2
10 4
x x
 
+25 +25
2
10 25 29
x x
  
 
2
5 29
x  
5 29
x   
5 29
x   
Ex: 4Solve by Completing the Square
0
11
8
2


 x
x
11
8
2


 x
x
+16 +16
5
16
8
2


 x
x
  5
4
2


x
  5
4 


x
5
4 


x
Ex 5:The coefficient of x2 must be
“1”
0
3
3
2 2


 x
x
0
2
3
2
3
2


 x
x
16
33
4
3
2








x
2
3
2
3
2

 x
x
4
3
2
2
3


2 9 9
16 16
3 3
2 2
x x
 
 
16
33
4
3



x
16
33
4
3



x
2 2 2 2
33
4
4
33
3


x
Ex 6: The coefficient of x2 must be
“1”
2
3 12 1 0
x x
  
2 1
4
3
x x
  
2
4 4
1
4
3
x x
  
 
 
2 11
2
3
x  
11
2
3
x   
11
2
3
x   3
3

33
2
3
x  
33
2
3
x  
6
3
6 33
3
x



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Math-9-Lesson-drtxtxrxrdrdtcydWelcome to Gboard clipboard, any text you copy will be saved here.Welcome to Gboard clipboard, any text you copy will be savedfdyfyfyf here.3.pdf

  • 1. Completing the Square • Objective: To complete a square for a quadratic equation and solve by completing the square
  • 2. Steps to complete the square • 1.) You will get an expression that looks like this: AX²+ BX • 2.) Our goal is to make a square such that we have (a + b)² = a² +2ab + b² • 3.) We take ½ of the X coefficient (Divide the number in front of the X by 2) • 4.) Then square that number
  • 3. • Take half of the coefficient of ‘x’ • Square it and add it To Complete the Square x2 + 6x 3 9 x2 + 6x + 9 = (x + 3)2
  • 4. Complete the square, and show what the perfect square is: x x 12 2  36 12 2   x x  2 6  x y y 14 2  49 14 2   y y  2 7  y y y 10 2  25 10 2   y y  2 5  y x x 5 2  4 25 5 2   x x 2 2 5        x
  • 5. To solve by completing the square • If a quadratic equation does not factor we can solve it by two different methods • 1.) Completing the Square (today’s lesson) • 2.) Quadratic Formula (Monday’s lesson)
  • 6. Steps to solve by completing the square 1.) If the quadratic does not factor, move the constant to the other side of the equation Ex: x²-4x -7 =0 x²-4x=7 2.) Work with the x²+ x side of the equation and complete the square by taking ½ of the coefficient of x and squaring Ex. x² -4x 4/2= 2²=4 3.) Add the number you got to complete the square to both sides of the equation Ex: x² -4x +4 = 7 +4 4.)Simplify your trinomial square Ex: (x-2)² =11 5.)Take the square root of both sides of the equation Ex: x-2 =±√11 6.) solve for x Ex: x=2±√11
  • 7. Ex 1: Solve by Completing the Square 2 6 16 0 x x    2 6 16 x x   +9 +9 2 6 9 25 x x      2 3 25 x     3 5 x    3 5 x   8 x  2 x  
  • 8. Ex 2: Solve by Completing the Square 2 2 5 0 x x    2 2 5 x x   +1 +1 2 2 1 6 x x      2 1 6 x     1 6 x    1 6 x  
  • 9. Ex 3: Solve by Completing the Square 2 10 4 0 x x    2 10 4 x x   +25 +25 2 10 25 29 x x      2 5 29 x   5 29 x    5 29 x   
  • 10. Ex: 4Solve by Completing the Square 0 11 8 2    x x 11 8 2    x x +16 +16 5 16 8 2    x x   5 4 2   x   5 4    x 5 4    x
  • 11. Ex 5:The coefficient of x2 must be “1” 0 3 3 2 2    x x 0 2 3 2 3 2    x x 16 33 4 3 2         x 2 3 2 3 2   x x 4 3 2 2 3   2 9 9 16 16 3 3 2 2 x x     16 33 4 3    x 16 33 4 3    x 2 2 2 2 33 4 4 33 3   x
  • 12. Ex 6: The coefficient of x2 must be “1” 2 3 12 1 0 x x    2 1 4 3 x x    2 4 4 1 4 3 x x        2 11 2 3 x   11 2 3 x    11 2 3 x   3 3  33 2 3 x  
  • 13. 33 2 3 x   6 3 6 33 3 x  