high school complex numbers

nevinsmith · 2011-12-30 18:58:40

Hi, this question was in the 2011 new zealand calculus scholarship paper,

couldn't do it, don't know the answer,

x^6+(2-k)x^4+(25-2k)x^2-25k=0

given c=2^0.5+i3^0.5 ^o.5 are sqr roots
is one of the solutions of x.

find the other 5 roots.

thanks

nevinsmith · 2011-12-30 20:41:03

heres a link link to the paper, it was reuploaded online

http://www.nzqa.govt.nz/assets/scholarship/2011/93202-qbk-2011.pdf

it's question 5b

Im going to be doing chemical engineering, hope the maths don't get too crazy in university
XD

thanks to everyone

bobbym · 2011-12-30 20:55:17

Hi;

You have to play a little with these. This will get you started.

Make the substitution

polynomial is now.

This factorization is tough but can be done with some trial and error.

Remember your substitution:

You now have 3 roots, that should start you off. Can you finish?

Bob · 2011-12-30 21:08:24

hi nevensmith

As the equation has real coefficients and only even powers of x, you may state that

are also solutions.

By formulating the quartic from those you should be able to deduce k and find the remaining roots.

Bob

Good morning bobbym

Last edited by Bob (2011-12-30 21:09:27)

nevinsmith · 2011-12-30 21:22:02

hi bobbym,

so they are,
-rt2+rt3i
-rt2-rt3i
rt2-rt3i
-rt2-rt3i
rt2-rt3i

with 2 of the roots being repeated?

nevinsmith · 2011-12-30 21:24:14

this question i got stuck at the part where you factorise out y-k

in suituations like this how did you factorise that?
any special methods or just lots of practise

bobbym · 2011-12-30 21:26:16

Hi Bob;

Sorry for the delay I was called away. Good morning!

Hi nevinsmith;

I have these two roots and you already had one.

That factorization took me about 30 minutes. Nothing but pure trial and error and hoping that the form looked like that.

Checking with mathematica, he gets it instantly, although mine is prettier. Sorry, I just like to factor polys, next time I will use the program so that you will get a faster answer.

any special methods or just lots of practise

For one thing contest problems always respond to general methods so I expanded out that cubic and began to play with this form

( y^2 + y + 25)( times by something ). I also trusted that the k would factor out neatly, after all there must be a solution to the problem. If this had been a real world problem I would not have even tried but contest problems are like schoolwork, very artificial. They are designed to illustrate a principle, they are clean, they are beautiful. Real world problems are a lot messier.

Also I have not solved the problem for you. You still have some work to do.

Bob · 2011-12-31 00:39:14

hi

I've now got all six roots, but, strangely, do not know k. Here's what I did:

If a is a solution the complement of a is also a soluition because the coefficients are real.

Also, as the expression is 'even' -a and - comp a must be solutions.

find that quartic.

Also

I'm going to use 'z' rather than 'x' for the equation here

but we are given x = root 2 and y = root 3 =>

Then we can work out the missing quadratic (by 'division') gives

So the other two roots are:

I had hoped that I would know k by now but not ???

Bob

Last edited by Bob (2011-12-31 00:41:09)

Bob · 2011-12-31 04:22:06

hi again

note:

are the same solutions as:

Just try squaring one of these.

As for the last two solutions, and not knowing k, I've come to the conclusion that k may be arbitarily chosen.

Bob

anonimnystefy · 2011-12-31 04:25:02

hi bob

another miracle.happy new year to you!!!

Bob · 2011-12-31 04:28:59

And a very Happy New Year to you.

As for the miracles, I think you may need to adjust your probabilities to take account of recent evidence.

Bob

anonimnystefy · 2011-12-31 04:37:53

hi bob

i can explain it very nicely.you see because it's a school holiday right now,the P(Stefy has work for school) is an infinitesimal and the only know infinitesimal is,as you know well,...0.which means that the probability of us meeting is greater than when the P(SHWFS)=pi^2/10.so in the local universe this is a pretty common event,but in the global universe this is a very rare event,which makes it tend towards being a miracle,so as longitude of the rest of the holiday approaches zero,the event of us meeting comes closer and closer to being a miracle,so we can consider it to be one (approximately).

Bob · 2011-12-31 04:45:05

Your 'miracle theory' is progressing.

Now you have approximate miracles!

Soon I expect there will be 'complex miracles'. That will come in handy for explaining to non-mathematicians how 'imaginary' numbers can, nevertheless, exist. We can just say, "It's a miracle!"

Bob

anonimnystefy · 2011-12-31 04:50:50

hi bob

complex miracles?

and i'm not sure they would buy it.let's suggest new terms for non-mathematicians.we could call them normal people or just people.

Bob · 2011-12-31 04:52:46

=> mathematicians are not normal or are unjust.

I think we could call them muggles.

Bob

Last edited by Bob (2011-12-31 04:54:29)

anonimnystefy · 2011-12-31 04:58:03

your logic has gotten a little rusty. it should be 'and are unjust' using de Morgan's laws.

muggles? what does that mean?

maybe we could call them mentally unchallenged people?

Last edited by anonimnystefy (2011-12-31 04:58:30)

Bob · 2011-12-31 05:08:19

your logic has gotten a little rusty. it should be 'and are unjust' using de Morgan's laws.

I stand corrected.

muggles? what does that mean?

http://en.wikipedia.org/wiki/Muggle

maybe we could call them mentally unchallenged people?

Now you are getting close. An interesting use of English.

Normally mentally challenged would mean "not very bright" but I like the idea of "mentally unchallenged" which I am taking to mean "meets a maths problem and gives up".

So this a good suggestion. With one small flaw. The expression takes longer to type than "non-mathematician" so where is the advantage?

How about abbreviating it to "MUP"?

B

anonimnystefy · 2011-12-31 05:12:58

hi B

i must ask you why you are now signing only with B and not Bob.

yes the abbreviation sounds nice.and it's an unintentional use of English,because i thought while the non-mathematicians are MU,mathematicians are MC (if you know what i mean... ).

Bob · 2011-12-31 05:18:31

I haven't changed my name. This is the longest sequence of posts I have ever made in one session and I'm just trying to reduce the finger wear.

If

But would you accept "MACHs"

I'm going to have to log off or I'll start a family argument. We have visiters.

Bob

anonimnystefy · 2011-12-31 05:19:50

hi bob

"MACHs"???

Bob · 2011-12-31 05:22:02

Short for

MAthematically CHallenged.

B

anonimnystefy · 2011-12-31 05:24:31

i wasn't thinking of mathematically challenged,but something else...never mind.

Math Is Fun Forum

#1 2011-12-30 18:58:40

high school complex numbers

#2 2011-12-30 20:41:03

Re: high school complex numbers

#3 2011-12-30 20:55:17

Re: high school complex numbers

#4 2011-12-30 21:08:24

Re: high school complex numbers

#5 2011-12-30 21:22:02

Re: high school complex numbers

#6 2011-12-30 21:24:14

Re: high school complex numbers

#7 2011-12-30 21:26:16

Re: high school complex numbers

#8 2011-12-31 00:39:14

Re: high school complex numbers

#9 2011-12-31 04:22:06

Re: high school complex numbers

#10 2011-12-31 04:25:02

Re: high school complex numbers

#11 2011-12-31 04:28:59

Re: high school complex numbers

#12 2011-12-31 04:37:53

Re: high school complex numbers

#13 2011-12-31 04:45:05

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#14 2011-12-31 04:50:50

Re: high school complex numbers

#15 2011-12-31 04:52:46

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#16 2011-12-31 04:58:03

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#17 2011-12-31 05:08:19

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#18 2011-12-31 05:12:58

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#19 2011-12-31 05:18:31

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#20 2011-12-31 05:19:50

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#21 2011-12-31 05:22:02

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#22 2011-12-31 05:24:31

Re: high school complex numbers

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