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wolfgirl

  • 3 years ago

Solve the equation. Express the solution as a radical in the simplest form.

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  1. jhonyy9
    • 3 years ago
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    where is ?

  2. wolfgirl
    • 3 years ago
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    |dw:1314895631260:dw|

  3. jhonyy9
    • 3 years ago
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    16(x+3)=x2 x2 -16x -48=0

  4. wolfgirl
    • 3 years ago
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    do you mean x^2?

  5. fiddlearound
    • 3 years ago
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    square both sides.

  6. jhonyy9
    • 3 years ago
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    yes

  7. wolfgirl
    • 3 years ago
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    oh ok

  8. fiddlearound
    • 3 years ago
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    Just make sure that whatever solutions you find are valid for the original question.

  9. wolfgirl
    • 3 years ago
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    ok

  10. wolfgirl
    • 3 years ago
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    (x-12)(x-4)

  11. fiddlearound
    • 3 years ago
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    how did you get that ?

  12. wolfgirl
    • 3 years ago
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    I have no idea, let me try it again

  13. fiddlearound
    • 3 years ago
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    It's : x^2-16x-48=0 not: x^2-16x+48=0

  14. wolfgirl
    • 3 years ago
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    ok

  15. wolfgirl
    • 3 years ago
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    I got tht

  16. wolfgirl
    • 3 years ago
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    did I get the answer right though?

  17. fiddlearound
    • 3 years ago
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    You got this right: x^2-16x+48 = (x-12)(x-4) But we need to solve this: x^2-16x-48=0 Probably easiest to just use the quadratic formula.

  18. wolfgirl
    • 3 years ago
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    ok i'll try that

  19. wolfgirl
    • 3 years ago
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    24,-40

  20. fiddlearound
    • 3 years ago
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    can you show how you got that ?

  21. fiddlearound
    • 3 years ago
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    \[ax^2+bx+c=0\] \[x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\] What values did you use for: a=? b=? c=?

  22. wolfgirl
    • 3 years ago
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    sorry my comp was dissconnected again

  23. fiddlearound
    • 3 years ago
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    ok - so are you ready to continue solving the above ?

  24. wolfgirl
    • 3 years ago
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    yeah, I got -4,-10

  25. fiddlearound
    • 3 years ago
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    what values did you use for a=? b=? c=? when using the quadratic formula ?

  26. wolfgirl
    • 3 years ago
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    a=1 b=-16 c=-48

  27. fiddlearound
    • 3 years ago
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    right. can you show your full calculation ?

  28. fiddlearound
    • 3 years ago
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    first, can you calculate this part ? b^2-4*a*c That's the part inside the square root.

  29. wolfgirl
    • 3 years ago
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    |dw:1314900092634:dw||dw:1314899842897:dw|

  30. wolfgirl
    • 3 years ago
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    the top drawing is supposed to be on bottom

  31. fiddlearound
    • 3 years ago
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    I'll start with this guy under the square root. He is a special guy and even has a name: The discriminant. \[\Delta=b^2-4ac\] \[\Delta=b^2-4ac=(-16)^2-4(1)(-48)=256+192=448\] so, \[x=\frac{-b \pm \sqrt{\Delta}}{2a}=\frac{-(-16) \pm \sqrt{448}}{2(1)}=\frac{16 \pm \sqrt{7*8^2}}{2}\] \[=\frac{16 \pm 8\sqrt{7}}{2}=8 \pm 4\sqrt{7}\]

  32. fiddlearound
    • 3 years ago
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    let me know if the above is clear.

  33. wolfgirl
    • 3 years ago
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    i'm trying to understand it

  34. wolfgirl
    • 3 years ago
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    I don't get it

  35. fiddlearound
    • 3 years ago
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    can you tell me where it becomes unclear ?

  36. fiddlearound
    • 3 years ago
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    you basically did the same thing in the drawings you posted - just the negative/positive signs got messed up a bit.

  37. wolfgirl
    • 3 years ago
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    oh I get it, thanks

  38. fiddlearound
    • 3 years ago
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    we're not done yet though ... so if anything above is unclear, let me know before we go on. :)

  39. wolfgirl
    • 3 years ago
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    ok, explain the third line

  40. fiddlearound
    • 3 years ago
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    do you mean the part where the number 8 suddenly pops out from under the square root ?

  41. wolfgirl
    • 3 years ago
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    7*8^2

  42. fiddlearound
    • 3 years ago
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    oh - this ? 448 = 7*64 = 7*8*8 = 7*8^2

  43. wolfgirl
    • 3 years ago
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    yeah

  44. fiddlearound
    • 3 years ago
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    the reason I wrote it that way, is because the question wants us to: "Express the solution as a radical in the simplest form" Which means that we need to take everything we can out from under the square root. \[\sqrt{448}=\sqrt{7*64}=\sqrt{7*8^2}=\sqrt{7}\sqrt{8^2}=\sqrt{7}*8=8\sqrt{7}\] Is that cleat now ?

  45. wolfgirl
    • 3 years ago
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    yeah

  46. fiddlearound
    • 3 years ago
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    ok but we're still not done ...

  47. fiddlearound
    • 3 years ago
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    The quadratic equation gave us two solutions for x: \[x=8+4\sqrt{7}\] or \[x=8-4\sqrt{7}\] Right ?

  48. wolfgirl
    • 3 years ago
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    yeah

  49. fiddlearound
    • 3 years ago
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    but if you look at the original question, which was: \[4\sqrt{x+3}=x\] Notice the there is a square root there, and (x+3) is inside the square root. Right ?

  50. wolfgirl
    • 3 years ago
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    yeah

  51. fiddlearound
    • 3 years ago
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    Can a negative number have a real square root ?

  52. wolfgirl
    • 3 years ago
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    no

  53. fiddlearound
    • 3 years ago
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    so that means that since (x+3) is inside the square root, we cannot allow (x+3) to be negative - right ?

  54. wolfgirl
    • 3 years ago
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    right

  55. fiddlearound
    • 3 years ago
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    so we need to check the two possible values we got for x from the quadratic equation, and check that they don't make the number (x+3) a negative number. One of the numbers is easy to check, try checking this one first: \[x=8+4\sqrt{7}\] When you plug the above value into: \[(x+3)\] Do you get a negative value, or not ?

  56. wolfgirl
    • 3 years ago
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    positive

  57. wolfgirl
    • 3 years ago
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    is tht the answer?

  58. fiddlearound
    • 3 years ago
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    yes it is.

  59. wolfgirl
    • 3 years ago
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    ok thanks :)

  60. Ishaan94
    • 3 years ago
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    \[16(x + 3) = x^2\] \[x^2 - 16x - 48= 0 \]

  61. Ishaan94
    • 3 years ago
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    I did squaring ....

  62. Ishaan94
    • 3 years ago
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    Now I will find Solutions

  63. Ishaan94
    • 3 years ago
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    \[x = 4(2 +\sqrt7)\] \[x = 4(2 - \sqrt7)\] http://www.wolframalpha.com/input/?i=x^2+-+16x+-+48+%3D+0

  64. Ishaan94
    • 3 years ago
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    Now when we put the solution in Original function

  65. Ishaan94
    • 3 years ago
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    \[4(\sqrt{x+3}) = x \]I mean try putting \(4(2 -\sqrt7)\)

  66. Ishaan94
    • 3 years ago
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    You will get - in root which is not possible that is why only one solution is possible

  67. fiddlearound
    • 3 years ago
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    yes, but plugging in that value into (x+3), it is not negative: (x+3) = (4*(2-sqrt(7)))+3 = 0.41...

  68. Ishaan94
    • 3 years ago
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    Yeah You are right ...I did -3 in 8 my mistake..hmm wolfram is wrong

  69. Ishaan94
    • 3 years ago
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    Let me check wolfram then

  70. Ishaan94
    • 3 years ago
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    http://www.wolframalpha.com/input/?i=4%28sqrt%283%2B8+-+4sqrt7%29%29%3D8+-+4sqrt7

  71. Ishaan94
    • 3 years ago
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    Check this out

  72. fiddlearound
    • 3 years ago
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    Checking: 8+4*sqrt(7) - Wolfram says true: http://www.wolframalpha.com/input/?i=4*sqrt%28%288%2B4*sqrt%287%29%29%2B3%29%3D%288%2B4*sqrt%287%29%29 Checking: 8-4*sqrt(7) - Wolfram says false: http://www.wolframalpha.com/input/?i=4*sqrt%28%288-4*sqrt%287%29%29%2B3%29%3D%288-4*sqrt%287%29%29

  73. Ishaan94
    • 3 years ago
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    Yup

  74. Ishaan94
    • 3 years ago
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    Something wrong with Wolfram I guess

  75. fiddlearound
    • 3 years ago
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    hhhmmmm... http://www.wolframalpha.com/input/?i=-4*sqrt%28%288-4*sqrt%287%29%29%2B3%29%3D%288-4*sqrt%287%29%29

  76. fiddlearound
    • 3 years ago
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    A result of squaring both sides ...

  77. fiddlearound
    • 3 years ago
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    wolfram is right.

  78. Ishaan94
    • 3 years ago
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    http://www.wolframalpha.com/input/?i=sqrt%2811+-4sqrt7%29*4 yeah you got it squaring removes the minus sign

  79. Ishaan94
    • 3 years ago
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    Lol I tried to prove that wolfram can be wrong time to time ...but it's always right..

  80. fiddlearound
    • 3 years ago
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    yeah, question is - does the teacher expect wolfgirl to take the two values from the quadratic equation and validate them in the original question.

  81. fiddlearound
    • 3 years ago
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    coz only one of them is valid.

  82. Ishaan94
    • 3 years ago
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    well I don't think teacher would expect that ...hmm maybe they allow calculators

  83. fiddlearound
    • 3 years ago
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    ok -- we can ask wolfgirl what they usually do in class. Maybe she's related to wolfram.

  84. Ishaan94
    • 3 years ago
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    Lol ... Good Job though ...really nice

  85. wolfgirl
    • 3 years ago
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    I'm not related to wolfram lol. But so far I am clueless as to what they do in class. I do Ashworth online school if that helps any

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