Ext 1 Maths. (1 Viewer)

nexusbrah

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Is it just me that is finding 3U hard or is everyone else? :newburn:
 

Lieutenant_21

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Is it just me that is finding 3U hard or is everyone else? :newburn:
It is significantly harder than the maths you have done so far but it is not difficult at all, you just need to change the way you think/study.
My advice to you is to try to learn WHY you got the question right and WHY you got it wrong and try to recognise a pattern or a structured approach to doing the questions.
What I did was doing questions from the Terry Lee textbook and never look at the solutions until I do the entire question then check my answer, if it is right I try to learn why it is right and if it wrong I try to learn why I got it wrong. If you develop your accuracy, speed will be developed overtime.
 

nexusbrah

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It is significantly harder than the maths you have done so far but it is not difficult at all, you just need to change the way you think/study.
My advice to you is to try to learn WHY you got the question right and WHY you got it wrong and try to recognise a pattern or a structured approach to doing the questions.
What I did was doing questions from the Terry Lee textbook and never look at the solutions until I do the entire question then check my answer, if it is right I try to learn why it is right and if it wrong I try to learn why I got it wrong. If you develop your accuracy, speed will be developed overtime.
I agree to what you said but im struggling with this

1361095775870.jpg
 

enoilgam

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Is it just me that is finding 3U hard or is everyone else? :newburn:
Well, when I did MX1 in Year 11 it definitely felt like a huge step up from Year 9-10 stuff (to be honest, even 2U felt like a step up). I think you will find that quite a few people are finding it difficult at this point - just try to keep with it and hang onto it until at least the end of preliminary.
 
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nexusbrah

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Well, when I did MX1 in Year 11 it definitely felt like a huge step up from Year 9-10 stuff (to be honest, even 2U felt like a step up). I think you will find that quite a few people are finding it difficult at this point - just try to keep with it and hang onto it until the end of preliminary.
2U is fine for me at the moment, I'm kind of understanding what im doing now with my work now
 

mac1996

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i reckon if you did advanced maths in your junior years, it would of helped a lot
in y11 (last yr) i found maths so hard during term 1/early term 2, i tried harder and ended up with first in y11
at times i was close to dropping ext 1, but at the end of the year i realised how stupid i was back then
never give up, at the end of the day hard work always pays off;)
 

Shadowdude

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It's supposed to be hard. It's not Extension maths for nothing.

Hopefully it'll inspire you to work hard and improve your mathematical skill.
 

nexusbrah

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i reckon if you did advanced maths in your junior years, it would of helped a lot
in y11 (last yr) i found maths so hard during term 1/early term 2, i tried harder and ended up with first in y11
at times i was close to dropping ext 1, but at the end of the year i realised how stupid i was back then
never give up, at the end of the day hard work always pays off;)
haha, im like that at the moment :p we're all gonna make it!
 

yasminee96

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I agree to what you said but im struggling with this

View attachment 27707
use the rules (memorise these :) ) : alpha + beta + gamma = -b/a
alphabeta + alphagamma + betagamma = c/a
alphabetagamma = -d/a

a. alpha + beta + gamma = - (-3)/1 = 3
b. alphabeta + betagamma + gammaalpha = 4/1 = 4
c. alphabetagamma = -2/1 = -2
d. 6 (alpha + beta + gamma) +5 = 6(3) + 5 = 23
e. 1/alphabeta + 1/betagamma + 1/alphagamma = (gamma + alpha + beta)/alphabetagamma = 3/-2
f. this is the 'tricky' one. alpha^2 + beta^2 + gamma^2 = (alpha + beta + gamma)^2 - 2(alphabeta + alphagamma + betagamma) = (3)^2 - 2(4) = 5

alpha, beta, gamma are roots
a, b, c, d are the coefficients of the polynomial ax^3 + bx^2 + cx +d
in this case you had 1x^3 + (-3)x^2 + 4x + 2
so a=1 , b=-3 , c=4 , d=2

pretty straightforward and youll get it with practice :)




EDIT: made a mistake. im stupid asdfghjkaljghfa
 
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nexusbrah

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use the rules (memorise these :) ) : alpha + beta + gamma = -b/a
alphabeta + alphagamma + betagamma = c/a
alphabetagamma = -d/a

a. alpha + beta + gamma = - (-3)/1 = 3
b. alphabeta + betagamma + gammaalpha = 4/1 = 4
c. alphabetagamma = -2/1 = -2
d. 6 (alpha + beta + gamma) +5 = 6(3) + 5 = 23
e. 1/alphabeta + 1/betagamma + 1/alphagamma = (gamma + alpha + beta)/alphabetagamma = 3/-2
f. this is the 'tricky' one. alpha^2 + beta^2 + gamma^2 = (alpha + beta + gamma)^2 - 2(alphabeta + alphagamma + betagamma) = (3)^2 - 2(4) = 5

alpha, beta, gamma are roots
a, b, c, d are the coefficients of the polynomial ax^3 + bx^2 + cx +d
in this case you had 1x^3 + (-3)x^2 + 4x + 2
so a=1 , b=-3 , c=4 , d=2

pretty straightforward and youll get it with practice :)




EDIT: made a mistake. im stupid asdfghjkaljghfa
yeah i picked it up, I realised what im meant to be doing
 

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