Solving Quadraticsby

Completing the Square&

Quadratic Formula

By: Jeffrey Bivin

Lake Zurich High School

jeff.bivin@lz95.org

Last Updated: October 24, 2007

X2 + 6x + 9

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2)3( x

X2 + 4x + 4

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4

2)2( x

X2 + 5x + 25/4

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225 )( x

X2 - 6x + 9

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2)3( x

Solve by Completing the Square

x2 + 10x + 8 = 0

(x2 + 10x ) = -8

(x2 + 10x + (5)2) = -8 + 25

(x + 5)2 = 17

(5)2 = 25

175 x175 x

51021

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Solve by Completing the Square

3x2 + 24x + 12 = 0

(x2 + 8x ) = -4

(x2 + 8x + (4)2) = -4 + 16

(x + 4)2 = 12

(4)2 = 16

124 x124 x

4821

x2 + 8x + 4 = 03

324 xJeff Bivin -- LZHS

Solve by Completing the Square

2x2 + 5x - 12 = 0

411

45 x

45

25

21

2

23

46 x

06252 xx

6)( 252 xx

16252

45

252 6))(( xx 16

252

45

1625

16962

45 )( x 16

121

411

45 )( x

4416 x

Jeff Bivin -- LZHS

Solve by Completing the Square

2x2 - 12x - 11 = 0

2583 x

3621

206 2

112 xx

2112 )6( xx

9))3(6( 21122 xx 93 2

218

2112)3( x 2

29

229)3( x

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Solve by Completing the Square

-5x2 + 12x + 19 = 0

5131

56 x

56

512

21

-505

195122 xx

519

5122 )( xx

2536

5192

56

5122 ))(( xx 5

362

56

2536

25952

56 )( x 25

131

5131

56 )( x

51316 x

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Solve by Completing the Square

5x2 - 30x + 45 = 0

3x

3621

50962 xx

9)6( 2 xx

99))3(6( 22 xx 93 2

0)3( 2 x0)3( x

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Solve by Completing the Square

5x2 - 30x + 75 = 0

63 ix

3621

501562 xx

15)6( 2 xx

915))3(6( 22 xx 93 2

6)3( 2 x6)3( ix

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Convert to vertex form

y = x2 + 10x + 8

y - 8 = (x2 + 10x )

y - 8 + 25 = (x2 + 10x + (5)2)

y + 17 = (x + 5)2 - 17

Axis of symmetry: x = -5

Vertex: (-5, -17)

y + 17 = (x2 + 10x + (5)2)

(5)2 = 25

x + 5 = 0

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Convert to vertex form

y = 5x2 - 30x + 46

y - 46 = 5(x2 - 6x )

y - 46 + 45 = 5(x2 - 6x + (-3)2)

y - 1 = 5(x - 3)2 + 1

Axis of symmetry: x = 3

Vertex: (3, 1)

y - 1 = 5(x2 - 6x + (-3)2)

5(-3)2 = 45

x - 3 = 0

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Solve by Completing the Square

ax2 + bx + c = 0

a

acb

abx 2

4

2

2

ab

ab

221

a02 a

cab xx

ac

ab xx )( 2

2

2

4

22

2 ))((ab

ac

ab

ab xx 2

2

4

2

2 ab

ab

2

2

4

22 )(

ab

ac

abx

2

2

4

4

a

acb

a

acb

abx 2

4

2

2

)(

a

acbbx 2

42 Jeff Bivin -- LZHS

Solve using the Quadratic Formula

3x2 + 7x - 4 = 0

a = 3b = 7c = -4

a

acbbx 2

42

32

)4(3477 2

x

6

48497 x

6977 x

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Solve using the Quadratic Formula

6x2 + 9x + 2 = 0

a = 6b = 9c = 2

a

acbbx 2

42

62

26499 2

x

12

48819 x

12339 x

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Solve using the Quadratic Formula5x2 - 8x + 1 = 0

a = 5b = -8c = 1

a

acbbx 2

42

52

154)8(8 2

x

10

20648 x

10448 x

101142 5

114 101128 x

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Solve using the Quadratic Formula6x2 - 17x - 14 = 0

a = 6b = -17c = -14

a

acbbx 2

42

62

1464)17(17 2

x

12

33628917 x

1262517 x

32

128 x

27

1242 x

122517 x

Jeff Bivin -- LZHS

Solve using the Quadratic Formulax2 + 6x + 9 = 0

a = 1b = 6c = 9

a

acbbx 2

42

12

91466 2

x

2

36366 x

206 x

26x 3

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Solve using the Quadratic Formula

3x2 + 7x + 5 = 0

a = 3b = 7c = 5

a

acbbx 2

42

32

53477 2

x

6

60497 x

6117 x

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6117 ix

Why do some quadratic equations have 2 real solutions

some have 1 real solution and some have two imaginary

solutions?

Now Consider

ax2 + bx + c = 0

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a

acbbx 2

42

If discriminant > 0, then 2 real solutions

If discriminant = 0, then 1 real solution

If discriminant < 0, then 2 imaginary solutions

890-11

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