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Firejay5 Group Title

Solve and show and explain work x^2 + bx + c = 0

  • one year ago
  • one year ago

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  1. LogicalApple Group Title
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    Try applying the quadratic equation to this one.

    • one year ago
  2. Firejay5 Group Title
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    @LogicalApple

    • one year ago
  3. LogicalApple Group Title
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    \[x = \frac{ -b \pm \sqrt{b^{2} - 4ac} }{ 2a }\]

    • one year ago
  4. Firejay5 Group Title
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    @BubbaMurphy Do you have a different way to do this problem???

    • one year ago
  5. BubbaMurphy Group Title
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    LogicApple's way is definitely the way to do it.

    • one year ago
  6. BubbaMurphy Group Title
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    in this case, a=1

    • one year ago
  7. Firejay5 Group Title
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    Is that really how yo would solve it

    • one year ago
  8. Firejay5 Group Title
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    would A, B, and C = 1

    • one year ago
  9. LogicalApple Group Title
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    Based on the equation you supplied, a = 1, b = b, and c = c.

    • one year ago
  10. LogicalApple Group Title
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    They are coefficients of whatever quadratic you are talking about. Did you have a quadratic in mind?

    • one year ago
  11. Firejay5 Group Title
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    what you mean

    • one year ago
  12. LogicalApple Group Title
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    \[6x^{2} - 9x + 12 = 0\] \[-5x^{2} + x - 10 = 0\] \[x^2 + x - 1 = 0\] These are some examples of quadratics, all set equal to 0.

    • one year ago
  13. Firejay5 Group Title
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    Yea I know that

    • one year ago
  14. LogicalApple Group Title
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    Oh, ok... so what is your question?

    • one year ago
  15. Firejay5 Group Title
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    We have to solve x^2 + bx + c = 0 for bonus and I have no idea what to do

    • one year ago
  16. LogicalApple Group Title
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    I suggested using the quadratic equation. The only difference between x^2 + bx + c = 0 and the quadratics that I mentioned is that the coefficients in your equation are only known by their variable letter. But that is ok, you can still apply the quadratic formula to them. The solutions to x^2 + bx + c = 0 is: \[\frac{ -b \pm \sqrt{b^{2} - 4c} }{ 2 }\] This is the same thing I wrote before, except I let a = 1 because the coefficient of x^2 in your equation is 1.

    • one year ago
  17. Firejay5 Group Title
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    We didn't learn that equation you gave me

    • one year ago
  18. LogicalApple Group Title
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    What about completing the square ?

    • one year ago
  19. Firejay5 Group Title
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    so is that the answer

    • one year ago
  20. LogicalApple Group Title
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    Yes, but you would arrive at the same answer if you complete the square. Both techniques are valid for solving quadratic equations.

    • one year ago
  21. Firejay5 Group Title
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    I have to show my work, so how would I show it

    • one year ago
  22. LogicalApple Group Title
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    Well if you're using the quadratic equation you could say "applying the quadratic equation". But if you wanted to complete the square then: |dw:1359081924838:dw|

    • one year ago
  23. Azteck Group Title
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    I think you can use sum and product of roots: \[\alpha +\beta= \frac{ -b }{ a }\] \[\alpha \beta=\frac{ c }{ a }\]

    • one year ago
  24. Firejay5 Group Title
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    b^2/4

    • one year ago
  25. LogicalApple Group Title
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    • one year ago
    1 Attachment
  26. LogicalApple Group Title
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    |dw:1359082578453:dw|

    • one year ago
  27. LogicalApple Group Title
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    (b/2)^2 = b^2/4

    • one year ago
  28. Firejay5 Group Title
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    After (x + b/2)^2 = b^2/4 - c. Find the square root of both.

    • one year ago
  29. LogicalApple Group Title
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    Yes

    • one year ago
  30. Firejay5 Group Title
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    I got x + b/2 = b/4 - c. Is that correct?

    • one year ago
  31. LogicalApple Group Title
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    no the square root of both sides gives: x + b/2 = ± sqrt(b^2/4 - c)

    • one year ago
  32. Firejay5 Group Title
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    what then

    • one year ago
  33. LogicalApple Group Title
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    All the steps are in the picture attached.

    • one year ago
    1 Attachment
  34. Firejay5 Group Title
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    I am sorry the last step in attachment is the answer

    • one year ago
  35. LogicalApple Group Title
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    Oh ok did you get it figured out ?

    • one year ago
  36. Firejay5 Group Title
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    so x = +/- sqrt b^2 - 4c /2 is the answer

    • one year ago
  37. LogicalApple Group Title
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    |dw:1359083804186:dw| yes

    • one year ago
  38. Firejay5 Group Title
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    Will you forgive me @LogicalApple ?

    • one year ago
  39. LogicalApple Group Title
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    What for? It's a learning experience

    • one year ago
  40. Firejay5 Group Title
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    I probably annoyed you a lot

    • one year ago
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