HSC 2012 MX2 Marathon (archive) (6 Viewers)

lolcakes52

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Re: 2012 HSC MX2 Marathon

Second part is done with Newton's method for x=k to find the approximation. I'm having trouble with the first part, possibly because I'm not sure about whether the next root is between k=1 and 1, greater than k, or less than zero.
 

barbernator

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Re: 2012 HSC MX2 Marathon

Second part is done with Newton's method for x=k to find the approximation. I'm having trouble with the first part, possibly because I'm not sure about whether the next root is between k=1 and 1, greater than k, or less than zero.
Well the sum of the roots is k, so if one root is between 0 and 1, one is between k-1 and k, then the next must be between 0 and 1 as well to satisfy k
 

math man

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Re: 2012 HSC MX2 Marathon

a better question would be to deduce the smallest positive root plus the smallest negative root is positve
 

seanieg89

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Re: 2012 HSC MX2 Marathon

Here is a decent question:

The three roots of the complex polynomial:



all lie on the unit circle in the complex plane. Prove that the three roots of:



also lie on the unit circle.
.
 

RealiseNothing

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Re: 2012 HSC MX2 Marathon

Here is a decent question:

The three roots of the complex polynomial:



all lie on the unit circle in the complex plane. Prove that the three roots of:



also lie on the unit circle.
So far I've got:



y/n?
 

RealiseNothing

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Re: 2012 HSC MX2 Marathon

I don't think q can possibly be that, that would imply that the product of roots of p has modulus 3.
Then what have I done wrong? Have I assumed something that isn't true maybe?

Let the roots of p(x) be alpha, beta, and gamma. Then:









Since the roots lie on the unit circle and hence .

I then do this for 'b' and 'c' as well to obtain my expression for q(x).
 

seanieg89

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Re: 2012 HSC MX2 Marathon

From your second line to your third line of displayed equations.
 

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