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    Find the positive constants a and b such that x^4 + 9/x^4 = (x^2 - a/(x^2))^2 + b for all non zero values of x.

    Damn!!!!! help.
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    (Original post by Dines)
    Find the positive constants a and b such that x^4 + 9/x^4 = (x^2 - a/(x^2))^2 + b for all non zero values of x.

    Damn!!!!! help.
    Is that x^4 + \dfrac{9}{x^4} = \left(x^2 - \dfrac{a}{x^2} \right)^2 + b ?

    Expand and compare coefficients.
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    (Original post by Kvothe the arcane)
    Is that x^4 + \dfrac{9}{x^4} = \left(x^2 - \dfrac{a}{x^2} \right)^2 + b ?

    Expand and compare coefficients.
    You there? How to expand. Just give me a starting hint.
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    (Original post by Dines)
    You there? How to expand. Just give me a starting hint.
    Expand the bracket as you would normally. (\lambda-\mu)^2=\lambda^2-2 \lambda \mu + \mu^2.
    Except in this case, \lambda=x^2 and \mu=\dfrac{a}{x^2}.
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    (Original post by Dines)
    How to expand this
    If you know how to expand brackets normally, then this is no different, if not see below
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    if it makes it easier say that u=a/x^2
    So then you would need to expand (x^2 -y)
    Then you can just substitute in the value of u later on
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    (Original post by Kvothe the arcane)
    Expand the bracket as you would normally. (\lambda-\mu)^2=\lambda^2-2 \lambda \mu + \mu^2.
    Except in this case, \lambda=x^2 and \mu=\dfrac{a}{x^2}.
    Wouldn't that be
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     \dfrac {a^2}{x^4}?
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    (Original post by Kvothe the arcane)
    Expand the bracket as you would normally. (\lambda-\mu)^2=\lambda^2-2 \lambda \mu + \mu^2.
    Except in this case, \lambda=x^2 and \mu=\dfrac{a}{x^2}.
    Mu should be a/x^2
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    (Original post by thefatone)
    Wouldn't that be
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     \dfrac {a^2}{x^4}?
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    \displaystyle \left(\dfrac{a}{x^2} \right)^2=\dfrac{a^2}{x^4} yes.
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    (Original post by Kvothe the arcane)
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    \displaystyle \left(\dfrac{a}{x^2} \right)^2=\dfrac{a^2}{x^4} yes.
    lol you edit mine aswell xD

    guess i should put in a spoiler xD
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    I gained the answer.
    a=3
    b=6

    Thank you.
 
 
 
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