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Dines
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#1
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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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Kvothe the Arcane
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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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Dines
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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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Kvothe the Arcane
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#4
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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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KaylaB
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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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thefatone
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#6
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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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KaylaB
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#7
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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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Kvothe the Arcane
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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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thefatone
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#9
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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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Dines
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#10
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I gained the answer.
a=3
b=6

Thank you.
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