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Hmm.. so 1. the product of roots must be +6 no matter whether the constant is +6 or -6 2. if the leading term doesn't have a magnitude of 1, the product of roots will still be +6 right? The HSC syllabus tells you a different thing. Stupid HSC :(

I think they didn't prove it by induction... Within their attempt at induction, they did something that Rorix said. But they didn't mean it as the reason why s_1 = 0 , s2 = -2p , s_3 = -3q

oh yes, haha... stupid me but.. we can change the definition of the symbols "6" and "11". 6 = 0 and 11 = 0 ; )

I think that's it, haha.. how easy. by the way, if someone knows, "although a few did prove the result required within their attempt at induction without realising that they had done so." will they get the marks?

I got all three (there are only three right?) the flower is the less common.

hmm I wasn't taking about English, but subjects like maths or science. >>(yes, they broke the code of the marking centre) << Do you mean there's a code which prohibits them from telling marking criteria / policy?

Thanks Archman. Do you know of any other way though? The marking guidelines says: correct solution -> 2 marks recognition that a^3 = -pa - q in this context or equivalent -> 1 mark (it's out of 2 marks) The comment says: A number attempted a proof by induction but were ultimately...

Oh, Rorix's rollercoaster doesn't look like a rolercoaster now. And... I just happen to think about this during one of my temporary insanity sessions.. Suppose someone said: "anyone who submits exams with text messaging acronyms gets an instant ZERO" I would disagree with him/her.

Let a, b and c be the roots of x^3 + px + q = 0, and define s_n by s_n = a^n + b^n + c^n , for n = 1, 2, 3, ... so, s_1 = 0 , s2 = -2p , s_3 = -3q Prove that for n > 3, s_n = - p s_(n-2) - q s_(n-3) This is from 2003 HSC BTW. Thanks.

in HSC exams will they tell you the meaning of some words which are not very commonly used throughout the course? (like frustum for example) EDIT: I realise that an understanding of the word frustum is not important in this question.

it wont work ^_^ "then (x-2)(x-3) divides into P(x) perfectly" : true "So you sub in the product of those roots randomly" not true you need to sub in (2-2)(3-3), i.e. you need to sub in two x values, one is 2 and the other is 3. the third one is.. whatever you want.. actually any two of...

ok.. he's wrong.. but how do you do part (iii) in post #10? can we do that kind of question?

hello hello hello.. this is a "what TEXTBOOK do you use?!" thread this is a "what TEXTBOOK do you use?!" thread this is a "what TEXTBOOK do you use?!" thread ;)

yea.. I dont think it works, hehe... (cubic root)^3 + (cubic root)^2 + (cubic root) + constant cant easily be made to look like a cubic. but something like (cubic root)^3 + (cubic root) + constant can.

I personally don't think you can sub @$ etc. Do it like ngai suggested somewhere down this thread. The approaches I was thinking of are 1. Find an equation whose roots are @$, @%, $%, call them y_1, y_2, y_3. then from that equation you can find y_1^3 + y_2^3 + y_3^3 How do we find...

by calculator, this isn't correct.

have you written down the complete Q (on BoS)? including all the lead-ups and part you don't think are really relevant? there might be some conditions for the roots which are given on previous parts...

Hmm... from the evidence you've been given so far, is it suffice to say that I'm going crazy? How much does it cost to go to a psychologist? BTW, be careful if you're planning to do 4U.. that course is cursed. Look at what our 4U geniuses have turned into... I managed to escape from the...

hmm.. probably.. I havent been feeling right recently too much study and no sleep I guess... now is play time :) well... I just think that everyone should be given the opportunity to see everything. At least if I give them a hint I've done my duty as a responsible person. ^_^

>> Of course, when you have to write down reasons for circle geometry putting ∪ in altn8 cegmnt R de same really wont do << It will, provided the marker is used to text messaging :) But it's still not a good idea though. EDIT: whoaa.. where did that quote come from??