1. Solving Quadratics by Completing the Square & Quadratic Formula
By: Jeffrey Bivin
Lake Zurich High School
Last Updated: October 24, 2007

2. X2 + 6x + 9 x 1 1 1 1 1 1 x x + 3 Now, complete the square 1 + 9 1 1
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3. X2 + 4x + 4 4 x 1 1 1 1 x x + 2 Now, complete the square 1 1 x + 2
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4. X2 + 5x + 25/4 x 1 1 .5 1 1 1 x x + 5/2 Now, complete the square 1
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5. X2 - 6x + 9 x 1 1 1 1 1 1 1 1 1 x - 3 x 1 + 9 1 1 turn 1 square over
turn 2 squares over
turn 2 squares over
turn 3 squares over
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6. Solve by Completing the Square
x2 + 10x + 8 = 0
(x2 + 10x ) = -8
(x2 + 10x + (5)2) =
(5)2 = 25
(x + 5)2 = 17
Jeff Bivin -- LZHS

7. Solve by Completing the Square
3x2 + 24x + 12 = 0 3
x2 + 8x + 4 = 0
(x2 + 8x ) = -4
(x2 + 8x + (4)2) =
(4)2 = 16
(x + 4)2 = 12
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8. Solve by Completing the Square
2x2 + 5x - 12 = 0 2
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9. Solve by Completing the Square
2x2 - 12x - 11 = 0 2
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10. Solve by Completing the Square
-5x2 + 12x + 19 = 0 -5
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11. Solve by Completing the Square
5x2 - 30x + 45 = 0 5
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12. Solve by Completing the Square
5x2 - 30x + 75 = 0 5
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13. Convert to vertex form y = x2 + 10x + 8 y - 8 = (x2 + 10x )
(5)2 = 25
y = (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)
Jeff Bivin -- LZHS

14. Convert to vertex form y = 5x2 - 30x + 46 y - 46 = 5(x2 - 6x )
5(-3)2 = 45
y = 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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16. Solve by Completing the Square
ax2 + bx + c = 0 a
The
Quadratic
Formula
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17. Solve using the Quadratic Formula
3x2 + 7x - 4 = 0
a = 3
b = 7
c = -4
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18. Solve using the Quadratic Formula
6x2 + 9x + 2 = 0
a = 6
b = 9
c = 2
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19. Solve using the Quadratic Formula
5x2 - 8x + 1 = 0
a = 5
b = -8
c = 1
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20. Solve using the Quadratic Formula
6x2 - 17x - 14 = 0
a = 6
b = -17
c = -14
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21. Solve using the Quadratic Formula
x2 + 6x + 9 = 0
a = 1
b = 6
c = 9
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22. Solve using the Quadratic Formula
3x2 + 7x + 5 = 0
a = 3
b = 7
c = 5
Two
Imaginary
Solutions
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23. Why do some quadratic equations have 2 real solutions some have 1 real solution and some have two imaginary solutions?

24. Now Consider Discriminant ax2 + bx + c = 089
-11
If discriminant > 0, then 2 real solutions
If discriminant = 0, then 1 real solution
If discriminant < 0, then 2 imaginary solutions
Jeff Bivin -- LZHS

25. That's
All
Folks
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