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Firejay5

  • 2 years ago

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

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  1. LogicalApple
    • 2 years ago
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    Try applying the quadratic equation to this one.

  2. Firejay5
    • 2 years ago
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    @LogicalApple

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

  4. Firejay5
    • 2 years ago
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    @BubbaMurphy Do you have a different way to do this problem???

  5. BubbaMurphy
    • 2 years ago
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    LogicApple's way is definitely the way to do it.

  6. BubbaMurphy
    • 2 years ago
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    in this case, a=1

  7. Firejay5
    • 2 years ago
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    Is that really how yo would solve it

  8. Firejay5
    • 2 years ago
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    would A, B, and C = 1

  9. LogicalApple
    • 2 years ago
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    Based on the equation you supplied, a = 1, b = b, and c = c.

  10. LogicalApple
    • 2 years ago
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    They are coefficients of whatever quadratic you are talking about. Did you have a quadratic in mind?

  11. Firejay5
    • 2 years ago
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    what you mean

  12. LogicalApple
    • 2 years ago
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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.

  13. Firejay5
    • 2 years ago
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    Yea I know that

  14. LogicalApple
    • 2 years ago
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    Oh, ok... so what is your question?

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

  16. LogicalApple
    • 2 years ago
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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.

  17. Firejay5
    • 2 years ago
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    We didn't learn that equation you gave me

  18. LogicalApple
    • 2 years ago
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    What about completing the square ?

  19. Firejay5
    • 2 years ago
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    so is that the answer

  20. LogicalApple
    • 2 years ago
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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.

  21. Firejay5
    • 2 years ago
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    I have to show my work, so how would I show it

  22. LogicalApple
    • 2 years ago
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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|

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

  24. Firejay5
    • 2 years ago
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    b^2/4

  25. LogicalApple
    • 2 years ago
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    1 Attachment
  26. LogicalApple
    • 2 years ago
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    |dw:1359082578453:dw|

  27. LogicalApple
    • 2 years ago
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    (b/2)^2 = b^2/4

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

  29. LogicalApple
    • 2 years ago
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    Yes

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

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

  32. Firejay5
    • 2 years ago
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    what then

  33. LogicalApple
    • 2 years ago
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    All the steps are in the picture attached.

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  34. Firejay5
    • 2 years ago
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    I am sorry the last step in attachment is the answer

  35. LogicalApple
    • 2 years ago
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    Oh ok did you get it figured out ?

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

  37. LogicalApple
    • 2 years ago
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    |dw:1359083804186:dw| yes

  38. Firejay5
    • 2 years ago
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    Will you forgive me @LogicalApple ?

  39. LogicalApple
    • 2 years ago
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    What for? It's a learning experience

  40. Firejay5
    • 2 years ago
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    I probably annoyed you a lot

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