Mega Maths Question to test your knowledge (1 Viewer)

[tommykins]

Just let m = x + (1/x) and solve.

If you're going to give us questions, give us better challenging ones. I'm unsure whether you simply can't do the question?

 

[lolokay]

well, I've been self teaching for a while (mostly from doing questions on here, reading bits of stuff on the internet) - but not specifically for 4unit

 

[Real Madrid]

it's a cahllegne question...I don't know how to answer.

 

[tommykins]

it's a cahllegne question...I don't know how to answer.

All your questinos so far have been in the form am² + bm + c = 0

just realise that it's a quadratic and let m = x + (1/x) and use the quadratic formula to solve.

 

[Real Madrid]

But each is more confusing then the next!

Argh!

I always sub m, but I don't knwo when to expand it.

 

[shaon0]

cos5x
= Re(cos[x] + isin[x])⁵
= c⁵ - 10c³s² + 5cs⁴
substituting 1-c² for s² and expanding + simplifying gives:
= 16c⁵ - 20c³ + 5c

32x^5 - 40x^3 + 10x - sqrt3 = 0
let x = cos[a]
2cos[5a] = sqrt3
cos[5a] = sqrt3 /2 = +-cos[pi/12 +- 2npi)
a = +-pi/60 +-2/5 npi

Damn...i was so close. I knew the answer for x but didn't put it.
Better luck next time

 

[lolokay]

Damn...i was so close. I knew the answer for x but didn't put it.
Better luck next time

yeah lol. In your solution you were up to the last step.. and then started to go backwards

 

[bubblesss]

sheesh u guys have started learning 4 unit maths??????//omg maybe i'm the only one who hasn't started???????

 

[shaon0]

yeah lol. In your solution you were up to the last step.. and then started to go backwards

How did you get the sqrt(3)....i didn't get it.

 

[tommykins]

How did you get the sqrt(3)....i didn't get it.

Find cos(5x).

Hence find the solutions to
32x^5 - 40x^3 + 10x - sqrt3 = 0
It's basically a "use the previous result, easy answer here"

But each is more confusing then the next!

Argh!

I always sub m, but I don't knwo when to expand it.

Expand what? if you're given something like (x^2+3x+6)^2 + 3(x^2+3x+6) - 9 = 0, the last thing you want to do is expand. ( dont do the question, i pulled it out of my ass)

 

[lolokay]

lol I just realised I didn't even solve that question fully. I only solved for a.
x = cos[a] = cos[pi/60 + {0,1,2,3,4}2/5 pi] or something

 

[bubblesss]

But each is more confusing then the next!

Argh!

I always sub m, but I don't knwo when to expand it.

in that sort of question all u have to do is sub m for the big term or coefficient of x^2
eg - (x^2 - 5x)^2 + (x^2 - 5x) +6=0
u just sub x^2-5x = m
usually but not always, the term u have to sub is in brackets.

 

[alex.leon]

(cos(x)+i.sin(x))^5
Let c=cosx and s=sinx
(cos(x)+i.sin(x))^5=(c+s.i)^5
= c^5 + 5c^4*s.i - 10c^3s^2 - 10c^2*s^3i + 5c.s^4 + s^5i
cos(5x) = c^5 - 10c^3*s^2 + 5c.s^4
since; s^2=1-c^2
cos(5x) = c^5 - 10c3(1-c^2) + 5c (1-c^2)^2
= 11c^5 - 10c^3 + 5c(1-2c^2+c^4)
= 16c^5 - 20c^3 + 5c
2*cos(5x)=32c^5 - 40c^3 +10c
Let c=x.
2*cos(5x)=32x^5 - 40x^3 +10x
Let 2*cos(5x)= sqrt(3)
Thus, sqrt(3) = 32x^5 - 40x^3 +10x
32x^5 - 40x^3 + 10x - sqrt3 = 0

LOL, i don't think my solutions correct.

why did i EVER stumble across this thread?


ow...my head...

 

[shaon0]

why did i EVER stumble across this thread?


ow...my head...

Who knows?