5.7 Completing the Square Day 2 

1

December 08, 2008

Dec 2­10:29 PM

5‐7 Completing the SquareDay 2

Objectives: *Solve equations by completing the square... *Rewrite functions by completing the square...

...when a does not equal one!!!

5.7 Completing the Square Day 2 

2

December 08, 2008

Dec 3­8:06 PM

Review #1: Find the missing value to complete the square.

x2 ‐ 14x +

( )2 = ( )2 = 49

x2 ‐ 14x + 49

b2

‐142

5.7 Completing the Square Day 2 

3

December 08, 2008

Dec 3­8:12 PM

( )2 = 25

x2 + 10x = ‐ 13

x2 + 10x + 25 = ‐13 + 25

(x + 5)2 = 12

x + 5 = ±√12

x = ‐5 ± 2√3

Review #2: Solve x2 + 10x + 13 = 0 by completing the square.

Review #3: Now, re‐write the functionin vertex form and find the vertex.

(x + 5)2 = 12

(x + 5)2 ‐ 12 = 0

f(x) = (x + 5)2 ‐ 12

vertex (‐5, ‐12)

10

2

5.7 Completing the Square Day 2 

4

December 08, 2008

Dec 3­8:05 PM

When the quadratic term has a coefficient that is not 1, you can still solve by completing the square and re‐write the quadratic in vertex form.

Example #1: Solve y = 2x2 ‐ 8x ‐ 15 by completing the square and re‐write the function in vertex form to find the vertex.

y = 2x2 ‐ 8x ‐ 15 2(x ‐ 2)2 = 23

2x2 ‐ 8x ‐ 15 = 0 2(x ‐ 2)2 ‐ 23 = 0

2x2 ‐ 8x = 15 y = 2(x ‐ 2)2 ‐ 23

2(x2 ‐ 4x) = 15 vertex (2, ‐23)

2(x2 ‐ 4x + 4) = 15 + 8

2(x ‐ 2)2 = 23

(x ‐ 2)2 = 11.5

x ‐ 2 = ±√11.5

x = 2 ±√11.5

5.7 Completing the Square Day 2 

5

December 08, 2008

Dec 2­10:30 PM

Example #2: Solve y = 3x2 + 15x ‐ 27 by completing the square and re‐write the function in vertex form.

y = 3x2 + 15x ‐ 27

3x2 + 15x ‐ 27 = 0

3x2 + 15x = 27

3(x2 + 5x) = 27

3(x2 + 5x + 6.25) = 27 + 18.75

3(x + 2.5)2 = 45.75

(x + 2.5)2 = 15.25

x + 2.5 = ±√15.25

x = ‐2.5 ±√15.25

52(    )

2

 (2.5)2 6.25

(x + 2.5)2 = 15.25

(x + 2.5)2 ‐ 15.25 = 0

y = (x + 2.5)2 ‐ 15.25

vertex (‐2.5, ‐15.25)

5.7 Completing the Square Day 2 

6

December 08, 2008

Dec 3­8:29 PM

Homework:page 285

﴾22 ­ 27, 30, 32, 33, 37, 39, 69 ­ 71﴿