quite a bunch of questions (2 Viewers)

[shuning]

all done

 

[GUSSSSSSSSSSSSS]

do you want me to do them and give u solutions tomorrow?????


plus is that hornsby library thing for today or tomorrow?????

 

[Timothy.Siu]

p(x)=(x-a)^2 Qx
p'(x)=(x-a)^2Q'(x)+Q(x)2(x-a)
=0 when x=a

next one u just use conjugate roots theorem and u divide it then by the conjugates multiplied together and u'll get a quadratic

 

[shuning]

p(x)=(x-a)^2 Qx
p'(x)=(x-a)^2Q'(x)+Q(x)2(x-a)
=0 when x=a

next one u just use conjugate roots theorem and u divide it then by the conjugates multiplied together and u'll get a quadratic

i did that for the 1st 1, but i dont think i get a mark for that .... T.T

and did read my 2nd question LOL? i already worked the 1st part out.... but cant factorize it...

 

[youngminii]

Lol why wouldn't you get the mark for the first one? That's the actual proof.

 

[youngminii]

For the second one, Edit: Oh lol it's just quadratic factors, listen to Timothy
For the third one, you draw a triangle (using radius) and then use trig to find the value of the angles, which you then work out to be the arg etc.
For fourth.. Use mod-arg form to divide.. And then lol I dunno how to do the n power thing, I would just use process of elimination (lol )
For fifth, you use the fact that cos(-x) = cos(x) and that cos(2pi - x) = cos(x).

 

[Timothy.Siu]

i did that for the 1st 1, but i dont think i get a mark for that .... T.T

and did read my 2nd question LOL? i already worked the 1st part out.... but cant factorize it...

i did read...i was telling u how to factorise it. if u read.

(x^2-2root3x+4)(x^2+2root3x+4)

 

[lolokay]

you could solve 10) iii) by solving each quadratic, then using the sum of the roots 2 at a time = 0 for the original equation

once you group the pairs together it simplifies down to 4cos[pi/9]cos[5pi/9] + 4cos[5pi/9]cos[7pi/9] + 4 cos[7pi/9]cos[pi/9] + 3 = 0
which gives the desired result

 

[shuning]

i did read...i was telling u how to factorise it. if u read.

(x^2-2root3x+4)(x^2+2root3x+4)

oh just wondering u from syd boys? cuz ur location says syd and i know some1 from syd boys whos name is tim (09er)

but meh.... tim is such a common name... syd boys is a big school prob 100 tims in ur year LOL

 

[Timothy.Siu]

oh just wondering u from syd boys? cuz ur location says syd and i know some1 from syd boys whos name is tim (09er)

but meh.... tim is such a common name... syd boys is a big school prob 100 tims in ur year LOL

yeah, theres like 4 tims in my year.

for the last one, u cud just equate coefficients of x^2 and u'll get it

 

[shuning]

Timothy.Siu.... whatta azn chinese name. and i looked at ur profile .. syd boys LOL

so yea.... i know tim is a common name and azns are common in ur school too. so i guess there is like 50 chinese ppl with the name tim in ur school?

still i kinda think ur the tim i know xD

edit: lol we posted in the same time...

 

[Timothy.Siu]

nah i dont really know random people out of school

 

[shuning]

nah i dont really know random people out of school

you went to tutor somewhere in city during year 10?

those names rings a bell?

Amy Zhang
Kenneth Luu

edit: prob not u tho, cuz that tim kid like sucks at math...... i fully pwn him LOL back then

 

[Timothy.Siu]

you went to tutor somewhere in city during year 10?

those names rings a bell?

Amy Zhang
Kenneth Luu

edit: prob not u tho, cuz that tim kid like sucks at math...... i fully pwn him LOL back then

nope

 

[shuning]

For the second one, Edit: Oh lol it's just quadratic factors, listen to Timothy

the discriminate is negative....... 16- 4 x 16

 

[shuning]

For the second one, Edit: Oh lol it's just quadratic
For fourth.. Use mod-arg form to divide.. And then lol I dunno how to do the n power thing, I would just use process of elimination (lol )

u cant get an exact angle for the arguement.

 

[shuning]

you could solve 10) iii) by solving each quadratic, then using the sum of the roots 2 at a time = 0 for the original equation

once you group the pairs together it simplifies down to 4cos[pi/9]cos[5pi/9] + 4cos[5pi/9]cos[7pi/9] + 4 cos[7pi/9]cos[pi/9] + 3 = 0
which gives the desired result

i origionallly thought of that too..... but how did u arrange it? i arranged it in conjugate pairs but didnt get the answer........

 

[Drongoski]

(with so many earlier responses, I may have overlooked .. so if this is a repeat I apologise !)


For Q10 (iii):

Use the identity in Q10(ii): equating the coeffs of z^2 of both sides, we get:

(1 + 1 + 1 + 4cos pi/9 cos 5pi/9 + 4 cos pi/9 cos 7pi/9 + 4 cos 5pi/9 cos 7pi/9 ) = 0

[No need to fully expand the RHS of identity in Q10(ii); we merely worry ourselves with the term in z^2 ]

Hence 3 + 4 ( the reqd expression ) = 0
Hence the reqd result.

 

[lolokay]

i origionallly thought of that too..... but how did u arrange it? i arranged it in conjugate pairs but didnt get the answer........

it's easier to equate coefficients as timothy + drongoski pointed out

for mine, with the roots, you get things in the form n +- rt(n^2 - 1), so if a,b,c = cos[1,5,7 pi/9]
you can then group things like (a + rt(a² - 1) + a - rt(a²-1))(b + rt(b²-1) + b - rt(b²-1)) [3 like this] + (a + rt(a²))(a - rt²) [3 like this as well]
= 4ab + 4bc + 4ac + 3

but as you can tell, this is a bit longer + difficult to set out/explain on paper

 

[jet]

Hopefully this is right. Sorry I can't help any further - i have public speaking tonight.