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Help with 12 (definitely not 11) in my pic Watch

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    (Original post by madfish)
    I don't think so but you are very close... the answer is 1.14 to 3dp according to my book
    The correct answer is 11.
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    Just multiply each term by \sqrt 2
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    No idea why we are taking the 6th root of anything


    2x^\frac{3}{2} = 3x^{-\frac{3}{2}}

    x^3 = \frac{3}{2}


    Since

    \frac{3}{2} + \frac{3}{2} = 3
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    (Original post by TenOfThem)
    No idea why we are taking the 6th root of anything


    2x^\frac{3}{2} = 3x^{-\frac{3}{2}}

    x^3 = \frac{3}{2}


    Since

    \frac{3}{2} + \frac{3}{2} = 3
    Oops I was misled by the title of the thread!!
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    (Original post by TenOfThem)
    No idea why we are taking the 6th root of anything


    2x^\frac{3}{2} = 3x^{-\frac{3}{2}}

    x^3 = \frac{3}{2}


    Since

    \frac{3}{2} + \frac{3}{2} = 3
    but these terms are squared with a +5 on the end ?
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    (Original post by madfish)
    so we just take the 6th root of 3/2 ? how on earth we do that without a calculator?
    No, you should find that x^3 = 3/2

    This will give you an approx value of 1.14 (and you will need a calculator)
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    (Original post by madfish)
    but these terms are squared with a +5 on the end ?
    You're making it too complicated for yourself

    The question tells you that f(x) = (something horrible) + 5
    and asks you to solve f(x) = 5

    So you just need to solve (something horrible) = 0

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    (Original post by davros)
    You're making it too complicated for yourself

    The question tells you that f(x) = (something horrible) + 5
    and asks you to solve f(x) = 5

    So you just need to solve (something horrible) = 0

    May I ask how I got it wrong Davros, does the power of 2 on the outside also effect the coefficient? am I missing something?
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    (Original post by davros)
    You're making it too complicated for yourself

    The question tells you that f(x) = (something horrible) + 5
    and asks you to solve f(x) = 5

    So you just need to solve (something horrible) = 0

    lol, i love how you put that haha

    this was in a non calculator paper :l
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    (Original post by TenOfThem)
    No idea why we are taking the 6th root of anything


    2x^\frac{3}{2} = 3x^{-\frac{3}{2}}

    x^3 = \frac{3}{2}


    Since

    \frac{3}{2} + \frac{3}{2} = 3
    why do I get this?
    Name:  image.jpg
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    (Original post by Robbie242)
    May I ask how I got it wrong Davros, does the power of 2 on the outside also effect the coefficient? am I missing something?
    If you had squared it you would have had 3 terms


    Why would you square it

    (function)^2=0 means function=0
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    (Original post by madfish)
    why do I get this?
    Name:  image.jpg
Views: 84
Size:  88.2 KB<--- me right now
    Because you have 3 where you should have 3/2
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    (Original post by Robbie242)
    May I ask how I got it wrong Davros, does the power of 2 on the outside also effect the coefficient? am I missing something?
    When you set the bracket equal to 0, you're basically left with

    2x^{3/2} = 3x^{-3/2}

    Multiply both sides by x^{3/2}
    to get
    2x^3 = 3
    and you're virtually there!

    I haven't checked your working earlier; you probably just made a minor slip somewhere
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    (Original post by TenOfThem)
    Because you have 3 where you should have 3/2
    but it says squared
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    (Original post by TenOfThem)
    If you had squared it you would have had 3 terms


    Why would you square it

    (function)^2=0 means function=0
    Oh I see I'd be exactly like (x+3)^2=0 so (x+3)=0 I see what you mean, my just being an idiot, /shame
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    (Original post by Robbie242)
    May I ask how I got it wrong Davros, does the power of 2 on the outside also effect the coefficient? am I missing something?
    If you actually square

    (2x^\frac{3}{2} - 3x^{-\frac{3}{2}})^2 = 4x^3 - 12 + 9x^{-3}
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    (Original post by madfish)
    but it says squared
    Madfish I recently recalled this after TenOfThem kicked common sense in my brain. say with have a function f(x)=(x+3)^2 say f(x)=0
    0=(x+3)^2 therefore (x+3)=0 no need to expand and expanding gets you a longer solution
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    (Original post by Robbie242)
    Madfish I recently recalled this after TenOfThem kicked common sense in my brain. say with have a function f(x)=(x+3)^2 say f(x)=0
    0=(x+3)^2 therefore (x+3)=0 no need to expand and expanding gets you a longer solution
    ahh I see, tenofthem is such an ace at mathematics
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    (Original post by madfish)
    ahh I see, tenofthem is such an ace at mathematics
    Indeed she is, I hope to be as good as her one day :cool:
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    (Original post by TenOfThem)
    If you actually square

    (2x^\frac{3}{2} - 3x^{-\frac{3}{2}})^2 = 4x^3 - 12 + 9x^3
    Ah cheers for clearing that up, unlikely but could possibly come up in c2 so thanks!
 
 
 
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