Solving Quadratics by Completing the Square

1
Solving QuadraticsbyCompleting the
SquareQuadratic Formula
By Jeffrey Bivin Lake Zurich High
School jeff.bivin_at_lz95.org
Last Updated October 24, 2007
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X2 6x 9
x
1
1
1
1
1
1
x
x 3
Now, complete the square
1
9
1
1
x 3
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X2 4x 4
x
1
1
1
1
x
x 2
Now, complete the square
1
4
1
x 2
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X2 5x 25/4
x
1
1
1
1
1
.5
x
x 5/2
Now, complete the square
1
25/4
1
.5
x 5/2
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X2 - 6x 9
x
1
1
1
1
1
1
1
1
1
x - 3
x
1
9
1
1
turn 1 square over
x - 3
turn 1 square over
turn 2 squares over
turn 2 squares over
turn 3 squares over
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Solve by Completing the Square

• x2 10x 8 0

(x2 10x ) -8
(x2 10x (5)2) -8 25
(5)2 25
(x 5)2 17
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Solve by Completing the Square

• 3x2 24x 12 0

3
x2 8x 4 0
(x2 8x ) -4
(x2 8x (4)2) -4 16
(4)2 16
(x 4)2 12
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Solve by Completing the Square

• 2x2 5x - 12 0

2
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Solve by Completing the Square

• 2x2 - 12x - 11 0

2
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Solve by Completing the Square

• -5x2 12x 19 0

-5
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Solve by Completing the Square

• 5x2 - 30x 45 0

5
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Solve by Completing the Square

• 5x2 - 30x 75 0

5
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Convert to vertex form

• y x2 10x 8

y - 8 (x2 10x )
(5)2 25
y - 8 25 (x2 10x (5)2)
y 17 (x2 10x (5)2)
y 17 (x 5)2 - 17
x 5 0
Axis of symmetry x -5
Vertex (-5, -17)
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Convert to vertex form

• y 5x2 - 30x 46

y - 46 5(x2 - 6x )
5(-3)2 45
y - 46 45 5(x2 - 6x (-3)2)
y - 1 5(x2 - 6x (-3)2)
y - 1 5(x - 3)2 1
x - 3 0
Axis of symmetry x 3
Vertex (3, 1)
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Solve by Completing the Square

• ax2 bx c 0

a
The Quadratic Formula
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Solve using the Quadratic Formula

• 3x2 7x - 4 0

a 3
b 7
c -4
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Solve using the Quadratic Formula

• 6x2 9x 2 0

a 6
b 9
c 2
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Solve using the Quadratic Formula

• 5x2 - 8x 1 0

a 5
b -8
c 1
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Solve using the Quadratic Formula

• 6x2 - 17x - 14 0

a 6
b -17
c -14
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Solve using the Quadratic Formula

• x2 6x 9 0

a 1
b 6
c 9
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Solve using the Quadratic Formula

• 3x2 7x 5 0

a 3
b 7
c 5
Two Imaginary Solutions
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Why do some quadratic equations have 2 real
solutionssome have 1 real solution and some
have two imaginary solutions?
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Now Consider

• ax2 bx c 0

Discriminant
89
0
-11
If discriminant gt 0, then 2 real solutions
If discriminant 0, then 1 real solution
If discriminant lt 0, then 2 imaginary solutions
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That's All Folks
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