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anonymous

  • 5 years ago

\[p = \sqrt{2 - p}\]

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  1. anonymous
    • 5 years ago
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    could you better yet show me how to do them?

  2. anonymous
    • 5 years ago
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    p=sqrt(2-p) First, square both sides. p²=2-p Then, throw everything to one side. p²+p-2=0 Then, factor it. (p+2)(p-1)=0 p+2=0 or p-1=0 p=-2 or p=1

  3. anonymous
    • 5 years ago
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    this goes against every rule in algebra so far. my teacher gave every one F's because he thinks we cant learn -_-

  4. amistre64
    • 5 years ago
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    sqrt(4) does not equal -2

  5. anonymous
    • 5 years ago
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    I wasn't done yet.

  6. amistre64
    • 5 years ago
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    1 is good

  7. anonymous
    • 5 years ago
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    waiit so what is the other one?

  8. anonymous
    • 5 years ago
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    2

  9. anonymous
    • 5 years ago
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    p=-2 is wrong, since sqrt(2-p) can't become lower than 2.

  10. amistre64
    • 5 years ago
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    only 1; when squareing sides, you run the risk of getting a false answer and have to check it back into the original to see if it works :)

  11. anonymous
    • 5 years ago
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    so then it is any number greater than 2?

  12. amistre64
    • 5 years ago
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    sqrt(2-p) shouldnt go below zero; below 2 is fine :)

  13. amistre64
    • 5 years ago
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    sqrt(2--2) = sqrt(4) = 2 -2 <> 2

  14. anonymous
    • 5 years ago
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    ok... this is confusing

  15. amistre64
    • 5 years ago
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    what is confusing about it?

  16. anonymous
    • 5 years ago
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    just the way my teacher told us only to do it a way i dont remember and like coding learn 2 at once you get raped... :/

  17. amistre64
    • 5 years ago
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    there are only two numbers whose value equals their sqrts; do you know those numbers?

  18. anonymous
    • 5 years ago
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    for this one?

  19. amistre64
    • 5 years ago
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    for any number.... does 0 = sqrt(0) ?? does 1 = sqrt(1) ??

  20. anonymous
    • 5 years ago
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    no sqrt for 1 0 = \[\sqrt{0}\]

  21. amistre64
    • 5 years ago
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    1*1 = 1 ; therefore; sqrt(1) = 1 right?

  22. anonymous
    • 5 years ago
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    wouldnt it equal \[\sqrt{1}\]

  23. amistre64
    • 5 years ago
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    1 = 1*1 = sqrt(1) = 1/1 = .....

  24. amistre64
    • 5 years ago
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    p = sqrt(2-p) what should we do to solve this?

  25. anonymous
    • 5 years ago
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    i dont even know. :/

  26. amistre64
    • 5 years ago
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    staring at it wont help :) Try something, even if its wrong....

  27. anonymous
    • 5 years ago
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    give me a minute...

  28. anonymous
    • 5 years ago
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    \[-2^{?} +p ^{?}\]

  29. anonymous
    • 5 years ago
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    ohh the question marks should be sqrd

  30. amistre64
    • 5 years ago
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    its a good attempt; but how do we 'do the same thing' to the other side of the equals?

  31. amistre64
    • 5 years ago
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    how do we 'undo' a square root?

  32. anonymous
    • 5 years ago
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    square it

  33. amistre64
    • 5 years ago
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    yes; so lets try that :) to both sides

  34. amistre64
    • 5 years ago
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    p^2 = sqrt(2-p)^2 p^2 = 2-p

  35. anonymous
    • 5 years ago
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    4 + psqrd = 0?

  36. amistre64
    • 5 years ago
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    do you see that steps I did?

  37. anonymous
    • 5 years ago
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    yes

  38. amistre64
    • 5 years ago
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    when you square the squarroot; sign; it disappears; that is the only thing that happens to that part....

  39. anonymous
    • 5 years ago
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    ok

  40. amistre64
    • 5 years ago
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    sqrt(15)^2 = 15 sqrt(3)^2 = 3 sqrt(23654)^2 = 23654 see it?

  41. amistre64
    • 5 years ago
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    sqrt(2-p)^2 = 2-p right?

  42. anonymous
    • 5 years ago
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    yeah

  43. amistre64
    • 5 years ago
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    p^2 = 2-p ; what do you wanna do to this to solve for p?

  44. anonymous
    • 5 years ago
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    same side?

  45. amistre64
    • 5 years ago
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    ok...lets put it all to the same side.... p^2 +p -2 = 2-2-p+p p^2 +p -2 = 0

  46. anonymous
    • 5 years ago
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    ok

  47. amistre64
    • 5 years ago
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    how do we 'unFOIL' this ?

  48. anonymous
    • 5 years ago
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    huhm flutter if i know.

  49. amistre64
    • 5 years ago
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    please dont use that language here

  50. anonymous
    • 5 years ago
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    kk

  51. anonymous
    • 5 years ago
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    i never understood the foil. my teacher writes it on the bored and says his thought erases then goes to the next prob

  52. amistre64
    • 5 years ago
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    FOIL is a term that is used to teach people how to multiply stuff inside of paranthesis

  53. anonymous
    • 5 years ago
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    ok

  54. amistre64
    • 5 years ago
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    they teach it in hopes that it will mean something later on; but really its a crutch that keeps you from learning whats really happening to a problem

  55. amistre64
    • 5 years ago
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    The distributive property is all that we need; do you remember the distributive property?

  56. anonymous
    • 5 years ago
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    i ask for help bet he says a smart alic reply to me... he needs a different career

  57. anonymous
    • 5 years ago
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    yeah

  58. amistre64
    • 5 years ago
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    How do we use the distributive property then?

  59. anonymous
    • 5 years ago
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    times outside the ( ) by the inside of the ( )

  60. amistre64
    • 5 years ago
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    very good :) We can do that as many times as it takes.... watch: Do you agree that 8*3 = 24?

  61. anonymous
    • 5 years ago
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    yep

  62. amistre64
    • 5 years ago
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    then 3(5+3) should also equal 24 right? 3(5) + 3(3) = 24 15 + 9 = 24 24 = 24

  63. anonymous
    • 5 years ago
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    ok

  64. amistre64
    • 5 years ago
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    and... (1+2)(5+3) should also equal 24..like this: (1+2)(5) + (1+2)(3) = 24 5(1) + 5(2) + 3(1) + 3(2) = 24 5 + 10 + 3 + 6 = 24 15 + 9 = 24 24 = 24 that is what they are calling FOIL..... its just using the distributive property twice

  65. anonymous
    • 5 years ago
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    ok so foil = complete distribution?

  66. amistre64
    • 5 years ago
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    yes :)

  67. anonymous
    • 5 years ago
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    ok.. u got a mic or something for pc? this would be easier...

  68. amistre64
    • 5 years ago
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    nope; using college computers..no sound no video no nothing but a keyboard

  69. anonymous
    • 5 years ago
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    darn... ok so please guide me through the problem im about to make a new post... itl be easier...

  70. amistre64
    • 5 years ago
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    ok...

  71. anonymous
    • 5 years ago
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    i posted it...

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