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Anguyennn

  • one year ago

how do solve 3x^3+2x^2-12x-8=0

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  1. zepdrix
    • one year ago
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    Ooo I think this one is going to work out nicely with grouping! :)

  2. Anguyennn
    • one year ago
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    how do you do that?

  3. Anguyennn
    • one year ago
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    yes that is what the question is asking me to do actually haha :)

  4. zepdrix
    • one year ago
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    Oh interesting c: hah

  5. zepdrix
    • one year ago
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    \[\large\rm \color{#DD4747}{3x^3-12x}+\color{#3366CF}{2x^2-8}=0\]Notice that the cubic term and the first degree term are both divisible by 3, while the squared term and constant term are both even, divisible by 2, so let's group them like this!

  6. zepdrix
    • one year ago
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    We want to do some factoring. What can we factor out of the red guys? What do they have in common?

  7. Anguyennn
    • one year ago
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    \[\large\rm \color{#DD4747}{3x^3-12x}+\color{#3366CF}{2x^2-8}=0\]

  8. Anguyennn
    • one year ago
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    all i see is that

  9. Anguyennn
    • one year ago
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    I don't see any equations

  10. zepdrix
    • one year ago
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    The equation didn't show up for you? Are you using Internet Explorer or some junk like that? :c lol

  11. Anguyennn
    • one year ago
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    I am using safari

  12. zepdrix
    • one year ago
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    oh boy :( hmm

  13. zepdrix
    • one year ago
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    Reload browser? :o Maybe the plugin just crashed for you or something

  14. Anguyennn
    • one year ago
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    ok please hold on ok

  15. Anguyennn
    • one year ago
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    oh yes i see it now!!

  16. zepdrix
    • one year ago
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    yay c:

  17. Anguyennn
    • one year ago
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    so it would be (3x+2) (x^2-4)

  18. zepdrix
    • one year ago
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    Oh, you did it all the way already XD haha nice!

  19. Anguyennn
    • one year ago
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    so would the final answer be x=2/3, -2, 2?

  20. zepdrix
    • one year ago
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    -2/3, ya? :) Gotta subtract the 2 over

  21. Anguyennn
    • one year ago
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    oh!!!

  22. Anguyennn
    • one year ago
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    but wait, what about x^3-8x^2=0

  23. zepdrix
    • one year ago
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    Oh this is another problem? I see :)

  24. Anguyennn
    • one year ago
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    this is another one

  25. Anguyennn
    • one year ago
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    yes

  26. zepdrix
    • one year ago
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    So they each have at least two x's, multiplying, ya? So we can pull an x^2 out of each.

  27. zepdrix
    • one year ago
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    \[\large\rm x^2(x-8)=0\]Something like that, ya? :d

  28. Anguyennn
    • one year ago
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    yes

  29. zepdrix
    • one year ago
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    Apply your `Zero-Factor Property` again :) \(\large\rm x^2=0\) and \(\large\rm x-8=0\) and solve for x in each case.

  30. Anguyennn
    • one year ago
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    I know one of them would be x=8

  31. Anguyennn
    • one year ago
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    but what about for the x^2? that is what i am confused about

  32. zepdrix
    • one year ago
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    \(\large\rm x^2=0\) Take square root, \(\large\rm \pm x=0\) Leading to, \(\large\rm x=0\)

  33. zepdrix
    • one year ago
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    square root of zero is... zero :) ya?

  34. Anguyennn
    • one year ago
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    yes

  35. Anguyennn
    • one year ago
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    but when you square root x it becomes 10

  36. zepdrix
    • one year ago
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    ?

  37. Anguyennn
    • one year ago
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    oh wait never mind, i am talking nonsense

  38. zepdrix
    • one year ago
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    x is being squared. we perform the `opposite` operation of squaring to remove the square. square root is the opposite of squaring. So when we apply the square root, they `undo` one another.

  39. Anguyennn
    • one year ago
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    wow thanks I understand it now!!

  40. zepdrix
    • one year ago
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    yay team \c:/

  41. Anguyennn
    • one year ago
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    are you able to help me with some other questions since you are here already :P

  42. Anguyennn
    • one year ago
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    you are a really great help btw thank you i appreciate it

  43. zepdrix
    • one year ago
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    uh sure +_+

  44. zepdrix
    • one year ago
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    you can post it, i gotta make some food real quick :D brb

  45. Anguyennn
    • one year ago
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  46. Anguyennn
    • one year ago
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  47. Anguyennn
    • one year ago
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    I am not sure how to do c, f, or g.

  48. zepdrix
    • one year ago
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    What did you get for b? Your profit function.

  49. Anguyennn
    • one year ago
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    I got 13.83n-102.4

  50. Anguyennn
    • one year ago
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    I showed you a picture as well

  51. Anguyennn
    • one year ago
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    the second document

  52. zepdrix
    • one year ago
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    oh lol :)

  53. Anguyennn
    • one year ago
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    yup :)

  54. zepdrix
    • one year ago
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    Since this is a `linear function`, \(\large\rm \text{rate of change = slope}\) So the slope of your P(n) function is your rate of change.

  55. Anguyennn
    • one year ago
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    how do you find out the slope? mx+b?

  56. zepdrix
    • one year ago
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    Yes, m :)

  57. zepdrix
    • one year ago
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    We can write the number like this if it makes things easier:\[\large\rm m=\frac{13.83}{1}\]Remember that slope is `rise over run`. So I would like two separate numerical values in my slope so I can interpret it. The 13.83 represents profit, the 1 represents 1 pizza sold. So this value represents the amount of profit made per pizza!

  58. Anguyennn
    • one year ago
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    hmm

  59. Anguyennn
    • one year ago
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    so would 1 be the y coordinate on the graph?

  60. zepdrix
    • one year ago
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    slope is `change in y` divided by `change in x`. So the bottom number is the x ( n in this case ).

  61. Anguyennn
    • one year ago
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    oh ok thank you so much for your help!

  62. Anguyennn
    • one year ago
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    until next time :)

  63. zepdrix
    • one year ago
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    `practical domain` You don't want to open a pizza shop and sell two pizzas every day. You'll end up losing money because of your operating costs! So you want your domain to start from the `break-even point`.

  64. Anguyennn
    • one year ago
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    which is going to be n=7.40?

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