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Brutus.

  • 2 years ago

Need some Assistance. (x-3)^2/(x+1)(x-2)>0 I'm supposed to solve it and express my answer in interval notation.

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  1. RaphaelFilgueiras
    • 2 years ago
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    |dw:1372837027496:dw|

  2. zzr0ck3r
    • 2 years ago
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    is this \[\frac{(x-3)^{2}}{(x+1)(x-2)}\]?

  3. Brutus.
    • 2 years ago
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    Yes. that is the equation.

  4. zzr0ck3r
    • 2 years ago
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    ok so the top is >0 for all x does not equal 3 right?

  5. Brutus.
    • 2 years ago
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    Right.

  6. zzr0ck3r
    • 2 years ago
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    so we need to make sure the bottom is > 0 right?

  7. Brutus.
    • 2 years ago
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    Yes.

  8. RaphaelFilgueiras
    • 2 years ago
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    |dw:1372837247856:dw|

  9. RaphaelFilgueiras
    • 2 years ago
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    i have forgot 3

  10. zzr0ck3r
    • 2 years ago
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    the bottom is a parabola that opens up, what are its x intercepts?

  11. Brutus.
    • 2 years ago
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    -1,2 Right?

  12. zzr0ck3r
    • 2 years ago
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    right

  13. zzr0ck3r
    • 2 years ago
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    so the bottom is greater than 0 when x<-1 and x>2, because it opens up and crosses y=0 at -1,2 with me?

  14. Brutus.
    • 2 years ago
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    So far yes.

  15. zzr0ck3r
    • 2 years ago
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    so we know the top is grater than 0 when x is not 3 we know the bottom is greater than 0 when x<-1 and x>2 so the whole thing is greater than 0 when x<-1 and x>2 and x does not equal 3

  16. zzr0ck3r
    • 2 years ago
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    (-infinity,-1)U(2,3)U(3,infinity)

  17. Brutus.
    • 2 years ago
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    Okay, I get it now.

  18. zzr0ck3r
    • 2 years ago
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    good deal:)

  19. Brutus.
    • 2 years ago
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    I have another quick question

  20. zzr0ck3r
    • 2 years ago
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    oki

  21. Brutus.
    • 2 years ago
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    I had another problem like this and I solved it and got x<-1. How would I write that in interval notation?

  22. zzr0ck3r
    • 2 years ago
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    (-infinity,-1)

  23. zzr0ck3r
    • 2 years ago
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    if it was <= (-infinity,-1]

  24. Brutus.
    • 2 years ago
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    The original problem was \[3(1-\frac{ 3 }{4 }x)>5-\frac{ 1 }{ 4 }x\]

  25. zzr0ck3r
    • 2 years ago
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    3-(9/4)x>5-(1/4)x -(8/4)x>5-3 2x<-2 x<-1 correct

  26. Brutus.
    • 2 years ago
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    \[x <-1\] I wrote the interval notation like so (-infinity,-1). But for some reason, my professor marked off points and circled interval notation.

  27. zzr0ck3r
    • 2 years ago
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    you sure it didn't say set builder?

  28. Brutus.
    • 2 years ago
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    The question said Solve the inequalities and express your answer in Interval notation.

  29. zzr0ck3r
    • 2 years ago
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    he/she is wrong http://www.regentsprep.org/Regents/math/ALGEBRA/AP1/IntervalNot.htm look at non ending interval note: we have ) not ] because we have < not <=

  30. zzr0ck3r
    • 2 years ago
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    just go talk to your teacher, I make mistakes grading all the time....

  31. Brutus.
    • 2 years ago
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    Okay, thank you!

  32. zzr0ck3r
    • 2 years ago
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    np

  33. Brutus.
    • 2 years ago
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    One more equation?

  34. Brutus.
    • 2 years ago
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    Inequality*

  35. zzr0ck3r
    • 2 years ago
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    sure

  36. Brutus.
    • 2 years ago
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    \[|\frac{x=1 }{ 4}|\ge4\]

  37. zzr0ck3r
    • 2 years ago
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    is that plus or minus?

  38. Brutus.
    • 2 years ago
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    I know your supposed to get two answers. I got one which was \[x \ge15\]

  39. Brutus.
    • 2 years ago
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    Plus. My bad

  40. Brutus.
    • 2 years ago
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    \[|\frac{ x+1 }{4 }|\]

  41. zzr0ck3r
    • 2 years ago
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    ok so this is strange and you may need to think about it a bit, but when you have |a|=b we get a=b and a=-b when you have |a|<b we get -b<x<b when you have |a|>b we get a<b and a>-b

  42. zzr0ck3r
    • 2 years ago
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    the very last and should be or

  43. zzr0ck3r
    • 2 years ago
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    *when you have |a|>b we get a<b or a>-b

  44. Brutus.
    • 2 years ago
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    Okay, So if I want the other answer I set the 4 to a negative answer?

  45. zzr0ck3r
    • 2 years ago
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    so you have abs((x+1)/4)>=4 so (x+1)/4>=4 gives the solutions you already have

  46. zzr0ck3r
    • 2 years ago
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    not exactly

  47. zzr0ck3r
    • 2 years ago
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    now we do (x+1)/4<=-4 x+1<=-16 x<=-17

  48. zzr0ck3r
    • 2 years ago
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    set it negative and flip the sign

  49. Brutus.
    • 2 years ago
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    Ah, that's what confused me.. Didn't know what to do with my sign

  50. zzr0ck3r
    • 2 years ago
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    you could have done -(x+1)/4<=4 but as a rule I just make the "non absolute value side" negative

  51. Brutus.
    • 2 years ago
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    Good rule! With interval notation, how would I write this?

  52. Brutus.
    • 2 years ago
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    I think its \[(-\infty,-17]U(\infty,15]\]

  53. zzr0ck3r
    • 2 years ago
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    (-infinity,-17]U[15,infinity)

  54. zzr0ck3r
    • 2 years ago
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    think of the number line, write it in the order it would fall on the number line

  55. Brutus.
    • 2 years ago
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    Oh, I see.

  56. zzr0ck3r
    • 2 years ago
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    and note that (infinity,5] does not make since, because if x were to be in there then infinity < x < 5

  57. zzr0ck3r
    • 2 years ago
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    <=5

  58. Brutus.
    • 2 years ago
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    You lost me there..

  59. Brutus.
    • 2 years ago
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    But, Thank you so much for the help.

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