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Easy Problem? Surely not binomial expansion... watch

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    Hey guys, my brother is doing GCSEs and asked me to help him with this problem his teacher set him. I did Furhter Maths last year, but looking at this it looked a bit like binomial expansion, which is way out of GCSE knowledge. But my maths is rusty, and i'm sure i've just missed the easy method. Any help? Thanks

    Given that x + 1/x = 5

    Find the value of x in:
    x^5 = 1/(x^5)
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    (Original post by Guaglio)
    Find the value of x in:
    x^5 = 1/(x^5)
    Take the 5th root of both sides.
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    (Original post by Kolya)
    Take the 5th root of both sides.
    and?
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    (Original post by Guaglio)
    and?
    That gives you x = 1/x. If you have x=1/x and x + 1/x = 5, it is easy to solve for x.
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    (Original post by Guaglio)
    Hey guys, my brother is doing GCSEs and asked me to help him with this problem his teacher set him. I did Furhter Maths last year, but looking at this it looked a bit like binomial expansion, which is way out of GCSE knowledge. But my maths is rusty, and i'm sure i've just missed the easy method. Any help? Thanks

    Given that x + 1/x = 5

    Find the value of x in:
    x^5 = 1/(x^5)
    it would help if you used latex.
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    (Original post by Guaglio)
    Hey guys, my brother is doing GCSEs and asked me to help him with this problem his teacher set him. I did Furhter Maths last year, but looking at this it looked a bit like binomial expansion, which is way out of GCSE knowledge. But my maths is rusty, and i'm sure i've just missed the easy method. Any help? Thanks

    Given that x + 1/x = 5

    Find the value of x in:
    x^5 = 1/(x^5)
    Do you mean x+\frac{1}{x}=5 which is what you have written, or \frac{1+x}{x}=5
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    This doesn't make sense to me.

    Thought you could just multiply both sides of x+\frac{1}{x} = 5 to get a quadratic. But that doesn't work numerically.

    Kolya's method gives 2.5 which doesn't seem to work.
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    I don't believe the question has been posted accurately.

    If x^5 = 1/x^5, then x^5 = +/-1 and so x = +/-1 (assuming x real). Which has no connection to "Given that x+1/x=5" (and indeed if x+1/x = 5, it is impossible that x^5 = 1/x^5).
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    (Original post by miml)
    This doesn't make sense to me.

    Thought you could just multiply both sides of to get a quadratic. But that doesn't work numerically.

    Kolya's method gives 2.5 which doesn't seem to work.
    Yeah, you're right. The question's wrong (or a trick). x+1/x=5 has two irrational solutions. x^5=1/x^5 has 10 complex solutions, only two of which are real (which are +-1). In other words, there are values of x which satisfy both equations.
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    Kolya: Did you mean "there are no values of x which satisfy both equations"?
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    LOL. Nvm
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    (Original post by Kolya)
    Yeah, you're right. The question's wrong (or a trick). x+1/x=5 has two irrational solutions. x^5=1/x^5 has 10 complex solutions, only two of which are real (which are +-1). In other words, there are values of x which satisfy both equations.
    This is a GCSE question.
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    this question is weird!!!
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    (Original post by DFranklin)
    Kolya: Did you mean "there are no values of x which satisfy both equations"?
    Yeah, typo. Sorry.

    Can you see where the flaw is in the method I posted right at the beginning, Dave? It's deceptively "legit", even though it doesn't actually work. i.e. the reasoning is:

    Note that if x is s.t. x^5 = 1/x^5, then x is also s.t. x = 1/x. (This loses solutions, but it still preserves truth.)
    Now, from the first equation, we have that 1/x = 5-x.
    Therefore, combining the two equations gives us x = 5-x.
    So x = 5/2 is a solution.
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    What's wrong is that if x = 5/2, then x does NOT equal 1/x. The two equations are inconsistent.

    (I kind of realise you knew that, but if they're inconsistent, you can't expect substituting one into the other to yield anything useful).
 
 
 
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