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Envii

  • 3 years ago

Please help.

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  1. Envii
    • 3 years ago
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    |dw:1362030447502:dw|

  2. Chelsea04
    • 3 years ago
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    what I think you should do is first multiply the (x-1) by (x^2-4)

  3. Chelsea04
    • 3 years ago
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    then square both sides, and then collect the like terms

  4. Envii
    • 3 years ago
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    How do I multiply (x-1) by (x^2-4)?

  5. Chelsea04
    • 3 years ago
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    expand the brackets

  6. Envii
    • 3 years ago
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    Is it x^3-x^2-4x+4?

  7. Chelsea04
    • 3 years ago
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    yea

  8. Envii
    • 3 years ago
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    Yay. :D Then how do I combine like terms?

  9. Chelsea04
    • 3 years ago
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    you have to square both sides first, then collect the like terms, then isolate x

  10. Chelsea04
    • 3 years ago
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    there might be an easier way, but i'm not sure

  11. Envii
    • 3 years ago
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    Square both sides? How?

  12. Chelsea04
    • 3 years ago
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    to get rid of the sqrt sign, you'll need to square both sides: \[(x^3-x^2-4x+4)^2=x^4-2x^3\]

  13. Envii
    • 3 years ago
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    Wait, wouldn't it be (x^4-2x^3)^2 on that side?

  14. Chelsea04
    • 3 years ago
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    no, because a sqrt and a squared cancels out. \[(\sqrt{a})^2=a\]because\[\sqrt{a}=a^\frac{ 1 }{ 2 }\]and\[(\sqrt{a})^2=a^{\frac{ 1 }{ 2 }*2}\]which is\[a^1=a\]

  15. Envii
    • 3 years ago
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    Oh I get it.

  16. Envii
    • 3 years ago
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    Do I have to do the (x^3-x^2-4x+4)^2 first?

  17. Chelsea04
    • 3 years ago
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    yes, now it's a very, very, very long process. Are you allowed to use a calculator for this part?

  18. Envii
    • 3 years ago
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    Yes I am.

  19. Chelsea04
    • 3 years ago
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    so type \[(x^3 −x^2 −4x+4)^2\] into the calculator to get an answer

  20. Envii
    • 3 years ago
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    How do I do that if I'm using the google calculator?

  21. Chelsea04
    • 3 years ago
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    I've never used a google calc. let me see

  22. Envii
    • 3 years ago
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    The calculator I normally use is broken, sorry. :(

  23. Chelsea04
    • 3 years ago
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    try use this site, http://www.wolframalpha.com/input/?i=%28x%5E3-x%5E2-4x%2B4%29%5E2 I've already typed in the equation, look for the words expanded form

  24. Envii
    • 3 years ago
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    Alright I got it. What next?

  25. Chelsea04
    • 3 years ago
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    subtract x^4 from both sides and add 2x^3 to get one side equal to 0

  26. Envii
    • 3 years ago
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    Alright, now I can do x^2-x^2 to get -, correct?

  27. Envii
    • 3 years ago
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    to get 0*

  28. Chelsea04
    • 3 years ago
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    how'd you get x^2?

  29. Chelsea04
    • 3 years ago
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    but, yes, use that idea to get all the terms on one side

  30. Envii
    • 3 years ago
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    x^6-x^4=x^2 and -2x^5+2x^3=x^2, right?

  31. Chelsea04
    • 3 years ago
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    no, you can't do that. they're not like terms. like terms are: x and 4x x^2 and 7x^2 not x^6 and x^4

  32. Chelsea04
    • 3 years ago
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    so what you should have gotten is: \[x^6-2x^5-7x^4+16x^3+8x^2-32x+16=x^4+2x^3\]

  33. whpalmer4
    • 3 years ago
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    Mind if I make a suggestion that will keep things simpler?

  34. Envii
    • 3 years ago
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    @whpalmer4 of course @Chelsea04 Wouldn't it be =x^4-2x^3?

  35. Chelsea04
    • 3 years ago
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    oh, yes sorry.

  36. whpalmer4
    • 3 years ago
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    Way back at the beginning, notice that \(x^2-4\) is the difference of two squares, and factors to \((x+2)(x-2)\) We can write the problem as \[(x-1) = \frac{\sqrt{x^4-2x^3}}{(x+2)(x-2)} = \frac{\sqrt{x^3(x-2)}}{(x+2)(x-2)} = \frac{x\sqrt{x(x-2)}}{(x+2)(x-2)}\]

  37. Chelsea04
    • 3 years ago
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    yea, but how does that help? I noticed it too, but didn't know what to do with it

  38. whpalmer4
    • 3 years ago
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    Now if we keep everything like that, and multiply both sides by the denominator, we get \[(x-1)(x+2)(x-2) = x\sqrt{x(x-2)}\]and after we square both sides, we just tack on exponents and get rid of the radical sign: \[(x-1)^2(x+2)^2(x-2)^2 = x^2x(x-2)\]and now we can cancel out a common term from both sides\[(x-1)^2(x+2)^2(x-2) = x^3\]

  39. Chelsea04
    • 3 years ago
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    then what?

  40. whpalmer4
    • 3 years ago
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    we marvel at how beautiful it is? :-)

  41. Chelsea04
    • 3 years ago
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    wouldn't you have to eventually expand it to solve for x? @whpalmer4 that's funny, lol :P

  42. whpalmer4
    • 3 years ago
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    Are we 100% certain this was the problem to be solved, and we're really supposed to find the roots? We've got a 5th order polynomial here...

  43. whpalmer4
    • 3 years ago
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    If we expand it out, we get \[x^5-8x^3+2x^2+12x-8=0\]

  44. whpalmer4
    • 3 years ago
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    Can't factor it, and there's no exact general solution for 5th order or higher polynomials, so Mathematica just finds numeric values for this one: \[\{\{x\to -2.53629\},\{x\to -1.59124\},\{x\to 0.8782\, -0.254473 i\},\]\[\{x\to 0.8782\, +0.254473 i\},\{x\to 2.37113\}\}\]

  45. Envii
    • 3 years ago
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    I am just so confused right now.

  46. Chelsea04
    • 3 years ago
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    what exactly does the question ask?

  47. Envii
    • 3 years ago
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    It just says solve.

  48. Chelsea04
    • 3 years ago
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    I guess you could've just typed the whole thing into the calculator

  49. whpalmer4
    • 3 years ago
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    Now here's a thought, what if the problem is really \[x-1 = \sqrt{\frac{x^4-2x^3}{x^2-4} }\]

  50. Chelsea04
    • 3 years ago
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    that'd make more sense

  51. whpalmer4
    • 3 years ago
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    Then we have \[(x-1)^2 = \frac{x^3(x-2)}{(x-2)(x+2)}\]

  52. Chelsea04
    • 3 years ago
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    is that the question @Envii ?

  53. whpalmer4
    • 3 years ago
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    Cancel out (x-2) and multiply both sides by (x+2) gives us \[(x-1)^2(x+2) = x^3\]Expand the left side \[(x^2-2x+1)(x+2) = x^3\]Expand again\[x^3+2x^2-2x^2-4x+x+2 = x^3\]

  54. whpalmer4
    • 3 years ago
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    @Envil, why don't you try solving that?

  55. whpalmer4
    • 3 years ago
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    This version is MUCH easier :-)

  56. Chelsea04
    • 3 years ago
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    YUP!!! MUCH MUCH EASIER!!!

  57. whpalmer4
    • 3 years ago
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    I did like the look of the other one in factored form, though :-)

  58. whpalmer4
    • 3 years ago
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    @Envil have you simplified and solved that equation yet?

  59. Envii
    • 3 years ago
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    Working on it. Sorry, I suck at math. xD

  60. whpalmer4
    • 3 years ago
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    Here's the equation again: \[x^3+2x^2-2x^2-4x+x+2 = x^3\]What do you have left after you collect like terms?

  61. Envii
    • 3 years ago
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    -4x+x+2=0

  62. whpalmer4
    • 3 years ago
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    Right. And what is -4x + x?

  63. whpalmer4
    • 3 years ago
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    Come on, you're really close to the answer!

  64. Envii
    • 3 years ago
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    -3x+2.

  65. whpalmer4
    • 3 years ago
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    \[-3x+2 = 0\]Add 3x to both sides, and then divide both sides by 3 to get x = ????

  66. Envii
    • 3 years ago
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    x=0.66?

  67. whpalmer4
    • 3 years ago
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    I'd keep it in exact form: x = 2/3

  68. whpalmer4
    • 3 years ago
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    But now we need to check it, because we squared both sides and that sometimes introduces fictitious solutions. x-1 = 2/3 -1 = -1/3 Uh oh, that's trouble, because we've got a square root on the other side of the equals sign, which is going to give us a positive number. If we expand it out, \[-\frac{1}3 = \sqrt{\frac{(\frac{2}{3})^4-2(\frac{2}{3})^3}{(\frac{2}{3})^2 -4}} = \sqrt{\frac{(\frac{16}{81})-2(\frac{8}{27})}{(\frac{4}{9}) -4}} = \sqrt{\frac{(\frac{16}{81})-(\frac{48}{81})}{(\frac{4}{9}) -(\frac{36}{9})}} = \sqrt{\frac{\frac{-32}{81}}{\frac{-32}{9}}}=\sqrt{\frac{1}9}\] So it isn't a solution after all.

  69. Envii
    • 3 years ago
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    You lost me. D:

  70. whpalmer4
    • 3 years ago
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    Well, did I lose you in the painful fraction arithmetic, or in the decision that the solution wasn't valid?

  71. whpalmer4
    • 3 years ago
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    After we plugged our "solution" back into the original equation and ground through the arithmetic, we ended up with this: \[-\frac{1}{3} = \sqrt{\frac{1}9}\]Unfortunately, that isn't true. Therefore, our solution isn't valid. There's a similar problem that does have working solutions: \[1-x = \sqrt{\frac{x^4-2x^3}{x^2-4}}\]We proceed the same way: \[(1-x)^2(x+2)(x-2) = x^3(x-2)\]Cancel common factors\[(1-x)^2(x+2)=x^3\]Expand\[(1-2x+x^2)(x+2)=x^3\]Expand again\[x+2-2x^2-4x+x^3+2x^2=x^3\]Collect like terms\[-3x+2=0\]\[x=\frac{2}3\] If we plug that back into its original equation: \[1-\frac{2}{3} = \sqrt{\frac{(\frac{2}{3})^4-2(\frac{2}{3})^3}{(\frac{2}{3})^2-4}} \]\[\frac{1}3 = \sqrt{\frac{1}{9}}\]and that is true...I wonder if that was the correct problem!

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