A.5 Completeing the Square 2010 September 15, 2010
A.5 Completing the Square
What you do when you are given a quadratic function in standard form and you need it to be in vertex form.
Objective: Complete the square.
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A.5 Completeing the Square 2010 September 15, 2010
Example:
2
x + 6x + 12 = 15
2
1) Get x and x alone on one side of the equation
2
x + 6x = 15 ­ 12
2
x + 6x = 3
2) Off to the side, compute: b = 6, so...
2
( )
6
2
2
(3)
2
( b2 )
9
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A.5 Completeing the Square 2010 September 15, 2010
3) Add this number to both sides of the equation
2
x + 6x + 9 = 3 + 9
*Remeber, you can do whatever you want to an equation as long as you do the same thing to both sides!
4) Factor on the left b
using , simplify on the 2
right
b 2
(x + )
2
2
x + 6x + 9 = 12
2
(x + 3) = 12
*If the term is positive, add it, if it's negative, subtract it.
2
5) Move the "loose" term back to the left side of the equation
(x + 3) ­ 12 = 0
6) Find the vertex
(x + 3) ­ 12 = 0
2
Vertex: (­3, ­12)
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A.5 Completeing the Square 2010 Example:
September 15, 2010
2
x + 3x ­ 8 = 0
b = 3
2
x + 3x = 8
2
( 32 )
9
4
2
x + 3x + = 8 + 9
4
Keep this a fraction
= 9
4
Make this a decimal
9
4
2
x + 3x + = 8 + 2.25
2
3
(x+ ) = 10.25
2
2
3
(x+ ) ­ 10.25 = 0
2
Vertex: ( ­
3
2
, ­10.25 )
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A.5 Completeing the Square 2010 Example:
September 15, 2010
2
x ­ 6x + 12 = 9
Answer
Vertex: (3, ­6)
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A.5 Completeing the Square 2010 Example:
September 15, 2010
2
3x ­ 18x + 8= 0
2
3(x ­ 6x ) + 8= 0
2
3(x ­ 6x ) = ­8
2
3(x ­ 6x + 9 ) = ­8 + 27
2
3(x ­ 3) = 19
2
3(x ­ 3) ­19 = 0
Answer
Vertex: (3, ­19)
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A.5 Completeing the Square 2010 Example:
September 15, 2010
2
4x ­ 16x + 9= 0
Vertex: (2, ­7)
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A.5 Completeing the Square 2010 September 15, 2010
Find the x­intercepts:
2
4x ­ 16x + 9= 0
2
4(x ­ 2) ­ 7= 0
2
4(x ­ 2) =7
2
(x ­ 2) = 7/4
(x ­ 2) = 7/4
+
Answer
x = 2 7/4
­
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A.5 Completeing the Square 2010 September 15, 2010
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