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Quadratic Factoring and Solving

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Show all of your work in a Word document and submit by the end of the module.
1-4: Solve by factoring:
1. 4x2 - 25 = 0
2. x2 - 12x + 36 = 0
3. x2 + 14x + 45 = 0
4. 6x2 - x - 15 = 0
5-7: Solve by completing the square.
5. x2 - 4x + 3 = 0
6. x2 + 5x - 1 = 0
7. 2x2 + 7x - 15 = 0
8-10: Solve by using the quadratic formula:
8. x2 + 9 = 0
9. x2 - 3x + 4 = 0
10. 2x2 + 3x = 6

11. Given the quadratic expression x2 + 4x + c, determine the value of "c" such that the equation has 2 real solutions. Repeat and find "c" such that the equation has 1 real solution. Finally, repeat such that the equation has no real solutions.

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Solution Summary

The solution gives detailed steps on factoring quadratic function, solving quadratic equations by using quadratic formula and by completing the square.

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1-4: Solve by factoring:
1. 4x2 - 25 = 0
Ans: 4x2-25=(2x-5)(2x+5)=0. So 2x-5=0 or 2x+5=0. Hence x=5/2 or x=-5/2

2. x2 - 12x + 36 = 0
Ans: x2 - 12x + 36=(x-6)(x-6)=0. So x-6=0. Hence x=6

3. x2 + 14x + 45 = 0
Ans: x2 + 14x + 45=(x+5)(x+9)=0. So x+5=0 or x+9=0. Hence x=-5 or x=-9

4. 6x2 - x - 15 = 0
Ans: 6x2 - x - 15=(2x+3)(3x-5)=0. So 2x+3=0 or 3x-5=0. Hence x=-3/2 or x=5/3

5-7: Solve by ...

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