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anonymous

  • one year ago

can anyone help??

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  1. anonymous
    • one year ago
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  2. anonymous
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  3. anonymous
    • one year ago
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    i think i just neeed to do the last 4 steps its asking at the bottom

  4. anonymous
    • one year ago
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    can anyone help or understand how to do that?

  5. xapproachesinfinity
    • one year ago
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    what you think u should be?

  6. xapproachesinfinity
    • one year ago
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    any ideas?

  7. xapproachesinfinity
    • one year ago
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    oh there two pages last 4 you are reffering to page 2 right?

  8. anonymous
    • one year ago
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    its all one problem

  9. xapproachesinfinity
    • one year ago
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    okay! first pages looks like an example to me

  10. anonymous
    • one year ago
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    i think i need to answer the 4 questions at the bottom of pg 2

  11. xapproachesinfinity
    • one year ago
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    anyways we are doing this \(\sqrt{(x^2-x)+7}=2(x^2-x)-1\) 1) is about the u

  12. xapproachesinfinity
    • one year ago
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    any idea?

  13. anonymous
    • one year ago
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    there is no u there, but in pg 11 it says let u= t^2-3t is that right?

  14. xapproachesinfinity
    • one year ago
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    it say what would u represent i may reading this incorrectly lol

  15. xapproachesinfinity
    • one year ago
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    they want you to suggest a u for that equation

  16. anonymous
    • one year ago
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    idk im very confused i dont see a U?? :S

  17. xapproachesinfinity
    • one year ago
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    no there is no u in the problem! what they are telling you is do substitution instead of solving for x we solve for a new variable u but we need first to make some kind of substitution if you looked at the first page you have they give you an example

  18. xapproachesinfinity
    • one year ago
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    i think you didn;t read what that page is saying otherwise you would be confused what u is lol

  19. xapproachesinfinity
    • one year ago
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    because that first page explains it all!

  20. anonymous
    • one year ago
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    i did read it.

  21. xapproachesinfinity
    • one year ago
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    lol! okay no problem so we need to do substitution to some variable u I'm speaking of this equation now \(\sqrt{(x^2-x)+7}=2(x^2-x)-1\)

  22. xapproachesinfinity
    • one year ago
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    this equation looks a mess for now that's why we do u to have only one good looking variable

  23. anonymous
    • one year ago
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    so we need to replace (x^2-x) with u right

  24. xapproachesinfinity
    • one year ago
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    so we can solve easily ah yes you are catching good going so far

  25. xapproachesinfinity
    • one year ago
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    \(\sqrt{u+7}=2u-1\) letting u=x^2-x yes

  26. xapproachesinfinity
    • one year ago
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    now we can solve for u before we go back to x

  27. xapproachesinfinity
    • one year ago
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    following?

  28. anonymous
    • one year ago
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    yea

  29. xapproachesinfinity
    • one year ago
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    \(\sqrt{u+7}=2u-1 \Longrightarrow u+7=(2u-1)^2\)

  30. xapproachesinfinity
    • one year ago
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    agree with this step or no?

  31. xapproachesinfinity
    • one year ago
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    i took square of both sides

  32. anonymous
    • one year ago
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    to get rid of the sqrt, yea

  33. xapproachesinfinity
    • one year ago
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    good! so now \(u+7=4u^2-4u+1 \Longrightarrow 4u^2-5u-6=0\)

  34. xapproachesinfinity
    • one year ago
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    we got a quadratic!

  35. xapproachesinfinity
    • one year ago
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    can we factor that!

  36. anonymous
    • one year ago
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    wait

  37. xapproachesinfinity
    • one year ago
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    yes! any questions

  38. anonymous
    • one year ago
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    to get 4u^2 -5u-6, you just did -u-7 from both sides right?

  39. xapproachesinfinity
    • one year ago
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    yes!

  40. anonymous
    • one year ago
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    and then we need to factr that to (4u+3)(u-2)?

  41. xapproachesinfinity
    • one year ago
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    now we factor to \((u-2)(4u+3)=0\)

  42. xapproachesinfinity
    • one year ago
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    yes excellent

  43. anonymous
    • one year ago
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    and use zpp?

  44. xapproachesinfinity
    • one year ago
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    solve for u what is zpp lol never heard this haha

  45. anonymous
    • one year ago
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    zero product property lol

  46. xapproachesinfinity
    • one year ago
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    i guess you mean u-2=0 or 4u+3=0 u=2 or u=-3/4

  47. anonymous
    • one year ago
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    yep

  48. xapproachesinfinity
    • one year ago
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    eh i see i never named anyhing it just makes sense to me with no name haha

  49. xapproachesinfinity
    • one year ago
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    now we solved for u but we still not finished for x

  50. xapproachesinfinity
    • one year ago
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    remember that we let u=x^2-x right

  51. anonymous
    • one year ago
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    so for the first question its u= *3/4 and u=2 right

  52. xapproachesinfinity
    • one year ago
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    so we have \(x^2-x=2 ~~or~~ x^2-x=-3/4\)

  53. xapproachesinfinity
    • one year ago
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    oh no first question to just answer u=x^2-x that is it

  54. xapproachesinfinity
    • one year ago
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    we are actually doing all those 4 question in one go

  55. anonymous
    • one year ago
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    ohh okay

  56. xapproachesinfinity
    • one year ago
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    once we finish you put it in order 1 put what we did 2 put what we did

  57. xapproachesinfinity
    • one year ago
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    here i went to answer those as one question lol they supposed to be one it is just that they want you to understand it part by part

  58. anonymous
    • one year ago
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    okay

  59. xapproachesinfinity
    • one year ago
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    2 we did we wrote the equATION WITH ONLY U

  60. anonymous
    • one year ago
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    gotcha

  61. xapproachesinfinity
    • one year ago
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    3 we did we solved for u u=2 u=-3/4

  62. xapproachesinfinity
    • one year ago
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    we are in 4 now :)

  63. xapproachesinfinity
    • one year ago
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    \(x^2-x=2 ~~or x^2-x=-3/4 \) we solve this now

  64. xapproachesinfinity
    • one year ago
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    two quadratic equations

  65. xapproachesinfinity
    • one year ago
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    \(x^2-x-2=0 ~~~or ~~~ 4x^2-4x+3=0\)

  66. xapproachesinfinity
    • one year ago
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    looks good to you?

  67. anonymous
    • one year ago
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    no lol

  68. xapproachesinfinity
    • one year ago
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    may be i wen too fast haha x^2-x=2 x^2-x-2=0 after subtracting 2 from both ssides

  69. anonymous
    • one year ago
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    how did you get 4x^2-4x+3

  70. xapproachesinfinity
    • one year ago
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    the other one x^2-x=-3/4 ===> 4(x^2-x)=-3 (multiplied 4 both sides) 4x^2-4x=-3 4x^2-4x+3=0 added 3 to both sides

  71. xapproachesinfinity
    • one year ago
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    clear now i guess haha

  72. xapproachesinfinity
    • one year ago
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    i just wanted to have a nice format i dislike fractions haha

  73. xapproachesinfinity
    • one year ago
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    good or nor good?

  74. anonymous
    • one year ago
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    i was just trying to understand it sorry, im a slow learner lol i got it tho

  75. anonymous
    • one year ago
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    i got it

  76. xapproachesinfinity
    • one year ago
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    hmm you are good learner as i see it :)

  77. anonymous
    • one year ago
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    aw thanks aha :))

  78. xapproachesinfinity
    • one year ago
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    we solve \(x^2-x-2=0 \Longrightarrow (x+1)(x-2)=0\)

  79. xapproachesinfinity
    • one year ago
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    this one 4x^2-4x+3=0 does not have real solutions

  80. xapproachesinfinity
    • one year ago
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    so we just solve (x+1)(x-2)=0

  81. xapproachesinfinity
    • one year ago
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    zpp lol

  82. anonymous
    • one year ago
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    x=1 and 2

  83. anonymous
    • one year ago
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    -1,+2

  84. xapproachesinfinity
    • one year ago
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    yes great job

  85. xapproachesinfinity
    • one year ago
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    that for step 4 and we are good!

  86. anonymous
    • one year ago
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    cool thanks so much

  87. xapproachesinfinity
    • one year ago
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    no problem

  88. anonymous
    • one year ago
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    you are a great teacher aha

  89. xapproachesinfinity
    • one year ago
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    not really :) i suck at some stuff haha

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