HSC 2013 MX2 Marathon (archive) (1 Viewer)

[RealiseNothing]

Re: HSC 2013 4U Marathon

-pi/4?

yep.

 

[RealiseNothing]

Re: HSC 2013 4U Marathon

-pi/4?

Try the old version of the question now, you should be able to get it if you got that one.

 

[twinklegal19]

Re: HSC 2013 4U Marathon

The quadratic has roots

0, since if all the coefficients are real, the roots must occur in conjugate pairs, therefore the arguments will be in opposite in sign?

edit: oh right it can also be pi as well

 

[HeroicPandas]

Re: HSC 2013 4U Marathon

Try the old version of the question now, you should be able to get it if you got that one.

But the new version has complex co-efficients and the old version has real co-efficients

 

[HeroicPandas]

Re: HSC 2013 4U Marathon

0, since if all the coefficients are real, the roots must occur in conjugate pairs, therefore the arguments will be in opposite in sign?

but he said in terms of a, b, c

Well I meant find the argument in terms of a, b, c.

 

[RealiseNothing]

Re: HSC 2013 4U Marathon

but he said in terms of a, b, c

You can do it that way too, both answers are correct really.

 

[deswa1]

Re: HSC 2013 4U Marathon

0, since if all the coefficients are real, the roots must occur in conjugate pairs, therefore the arguments will be in opposite in sign?

edit: oh right it can also be pi as well if both roots are real but one's negative and other's positive.

but he said in terms of a, b, c

0(a+b+c-3ab +2(a^2)b - c^3)=0 (that's in terms of a,b,c)

 

[RealiseNothing]

Re: HSC 2013 4U Marathon

0(a+b+c-3ab +2(a^2)b - c^3)=0 (that's in terms of a,b,c)

There's a much simpler expression lol.

 

[Carrotsticks]

Re: HSC 2013 4U Marathon

0, since if all the coefficients are real, the roots must occur in conjugate pairs, therefore the arguments will be in opposite in sign?

edit: oh right it can also be pi as well

How can it be pi as well?

 

[twinklegal19]

Re: HSC 2013 4U Marathon

How can it be pi as well?

if the roots are real but one's negative and the other positive?

but then technically it can be -pi too

 

[RealiseNothing]

Re: HSC 2013 4U Marathon

How can it be pi as well?

 

[deswa1]

Re: HSC 2013 4U Marathon

There's a much simpler expression lol.

It was a joke... Twink gave the answer as 0 and Herioc said it had to be in terms of a,b,c -> thus my answer

 

[SpiralFlex]

Re: HSC 2013 4U Marathon

I'm impressed that the advertisement of this thread worked to a degree.

 

[RealiseNothing]

Re: HSC 2013 4U Marathon

It was a joke... Twink gave the answer as 0 and Herioc said it had to be in terms of a,b,c -> thus my answer

oh lol, nvm then.

 

[bleakarcher]

Re: HSC 2013 4U Marathon

oh lol, nvm then.

realise, diggin' the sig bro.

 

[Sy123]

Re: HSC 2013 4U Marathon

damn I think I missed all the shortcuts

Nice work, alternatively while we can do it your way, do you see how you needed to find

And sum of them two at a time, then three at a time -> why not find a polynomial with roots 1/alpha 1/beta 1/gamma
And it actually isn't time consuming at all, in order to find a polynomial with those roots, you must 'reverse' the co-efficients
So a poly with roots 1/a 1/b 1/c
would be:

2x^3+3x^2+0x+1 = 0

So then you could find sum of them one at a time, two at a time and three by just using co-efficients instead of finding common denominator and going through that torturous process

And you could do the same for sum of squares and stuff as well.

 

[Immortalp00n]

Re: HSC 2013 4U Marathon

i came to this thread for d babes as per sy's sig
where them bad bitches at?

 

[Sy123]

Re: HSC 2013 4U Marathon

i came to this thread for d babes as per sy's sig
where them bad bitches at?

Err, not sure what you're talking about

Only maths here, if that's what you're talking about.

 

[RealiseNothing]

Re: HSC 2013 4U Marathon

i came to this thread for d babes as per sy's sig
where them bad bitches at?

Sorry about that.

 

[GoldyOrNugget]

Re: HSC 2013 4U Marathon

I was playing around with these sorts of problems on the bus the other day, but I ended up confusing myself and didn't get very far.

Say you had a polynomial with roots and you want to find a polynomial with roots (Sy's question would be a special case of this where ).

You can find an equation with these roots by replacing every instance of in the polynomial with (this is an interesting property to investigate/prove). In some cases, that equation can be algebraically manipulated into a polynomial without creating any new roots. This is why the trick of inverting coefficients works for -- it's a shortcut to replacing all terms with and multiplying out the resulting fractions.

When isn't monotonic (meaning is not a function), it still works; I'm pretty sure you can choose a maximal interval on its domain over which it's monotonic (probably the wrong terminology) and use that for the inverse, because the only necessity is that . Also, I'm not convinced that all equations generated in this manner can be manipulated into polynomials without creating new roots (for example, I had some trouble doing it with Sy's question, but I could have just messed up the algebra). Perhaps someone else will have more luck investigating this?