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**Yusuke00****Member**- Registered: 2013-11-19
- Posts: 43

Hello guys long time no see,got a new exercise:

On R[X] f=X^4-aX^3-aX+1

Proove that if |a|<1 then |x1|=|x2|=|x3|=|x4|=1;

Good luck.

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**Yusuke00****Member**- Registered: 2013-11-19
- Posts: 43

Ok guys a second problem

f(1)=8

f(2)=-5

f(3)=0

f(4)=35

f=?

The thing is i know how to solve the problem but i cannot really solve the system.

f(3)=0 -> x1=3 solution ->f = (X-3)(aX^2+bX+c)==>

==>f=aX^3+(b-3a)X^2+(c-3b)X-3c

so f(3)=0 ==>-3c=0 =>c=0

f(1)=8 ==> a+b=-4

f(2)=-5 ==>-4a-2b=-5==>4a+2b=5

f(4)=35 ==> 16a+4b=35

Can anyone solve the system or at least tell me where i'm wrong ? I'm getting very mad here.

*Last edited by Yusuke00 (2014-02-03 09:56:58)*

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

f(1)=9

f(2)=-5

f(3)=0

f(4)=35

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**Yusuke00****Member**- Registered: 2013-11-19
- Posts: 43

bobbym wrote:

Hi;

f(1)=9

f(2)=-5

f(3)=0

f(4)=35

Could you describe your solution step by step please? For f(1)=8 not 9 sorry.

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

After substitution into the general form you get.

Can you solve that or do you need a little more?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**Yusuke00****Member**- Registered: 2013-11-19
- Posts: 43

Ok i got those ecuations either.W/E i'm stupid.Thank you.

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Are you okay with getting the solutions? if you need help just post back.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**Yusuke00****Member**- Registered: 2013-11-19
- Posts: 43

Yes i had the same ecuations,but i've been tired and just couldn't focus yesterday.Got them now thank you.

By the way,any clues on 1st problem?

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,084

hi Yusuke00

I didn't post earlier because I do not understand what you mean by X1 X2 etc.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**Yusuke00****Member**- Registered: 2013-11-19
- Posts: 43

x1 are the solutions to the ecuation like...

f(x1)=0,f(x2)=0,f(x3)=0,f(x4)=0;

Sorry that's how i know to note them.

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,084

hi Yusuke00

That's what I thought you meant and that's where my difficulty began.

If you go to this page of the MIF site

http://www.mathsisfun.com/data/function … x)&func2=2

you will find a very useful function grapher. It even allows you to put in an 'a' and adjust its value using the slider.

This shows no roots at all.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**Yusuke00****Member**- Registered: 2013-11-19
- Posts: 43

For a>=1 there are real solutions.For a<1 there are not.That function grapher is for R not for C.

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**Yusuke00****Member**- Registered: 2013-11-19
- Posts: 43

But it's very helpful,i like it.Thank you very much ^_^.

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,084

On R[X] f=X^4-aX^3-aX+1

So what does R[X] mean ?

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**Yusuke00****Member**- Registered: 2013-11-19
- Posts: 43

It means that f(x) =y,y is part of R.

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