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radam:
Hey;

I'd appreciate some help with this question;

(from the previous subsection b is an imaginary root of the equation z^5-1=0 and b^4+b^3+b^2+b+1=0)
________________________________________
Q 13, PART B, MATH HL NOV 08

c) If u=b+b^4 and v= b^2+b^3, show that

i) u+v=uv=-1
ii) u-v= sqrt(5) , given that u-v>0
_______________________________________

for i) Proving that u+v=-1 is not difficult... but i have no idea how to prove that uv=-1, or for ii) how u-v is sqrt(5).

Help would be greatly appreciated :)

astarmathsandphysics:

--- Quote from: radam on April 23, 2009, 01:11:11 pm ---Hey;

I'd appreciate some help with this question;

(from the previous subsection b is an imaginary root of the equation z^5-1=0 and b^4+b^3+b^2+b+1=0)
________________________________________
Q 13, PART B, MATH HL NOV 08

c) If u=b+b^4 and v= b^2+b^3, show that

i) u+v=uv=-1
ii) u-v= sqrt(5) , given that u-v>0


_______________________________________

for i) Proving that u+v=-1 is not difficult... but i have no idea how to prove that uv=-1, or for ii) how u-v is sqrt(5).

Help would be greatly appreciated :)

--- End quote ---

i)uv=(b+b^4)(b^2+b^3)=b^3+b^4+b^6+b^7 now use b^5=1 ie B^6=b and B^7=b^2 so we have b^3+b^4+b +b^2=-1 too.

I will answer the second part as soon as I come back, but I have to go RIGHT NOW and teach some maths.

ii)(u-v)^2=u^2-2uv+v^2=(b+b^4)^2-2(b+b^4)(b^2+b^3)+(b^2+b^3)^2=b^2+2b^5+b^8-2b^3-2b^4-2b^6-2b^7_+b^4+2b^5+b^6
b^5=1 so we have b^2+2+b^3-2b^3-2b^4-2b-2b^2+b^4+b=(b^2+b^3-2b^3-2b^4-2b-2b^2+b^4+b)+4
=-(b^3+b^4+b +b^2)+4=5

radam:
Ah thanks!

I forgot that b^5=1;

I was thinking about the question with that bit of insight; and i was wondering if my method below is correct:

(u-v) is the same as sqroot (u-v)^2 (since u-v>0)

(U-v)=u^2-2uv+v^2= b^8+2b^5+b^2+2+b^4+2b^5+b^6

since uv=-1, and b^5=1

This gives b^4+b^3+b^2+b-6= 5

Hence  (u-v)^2= 5

u-v= sqrt(5)

The answer seems correct, but I'm wary of squaring as it normally adds complications to the matter... is there an easier and less time consuming way to do it?

Again thanks a lot for your help! With my math exam 2 weeks away, its a life-saver!

Clerk:
Hi. I'm a standard level physics student. Does anyone know if I have to know how to "Solve problems using the equation of state of an ideal gas". There are some past papers exercises, but I think that this part of the syllabus is now only for high level. Please help me!

astarmathsandphysics:

--- Quote from: radam on April 23, 2009, 02:04:40 pm ---Ah thanks!

I forgot that b^5=1;

I was thinking about the question with that bit of insight; and i was wondering if my method below is correct:

(u-v) is the same as sqroot (u-v)^2 (since u-v>0)

(U-v)=u^2-2uv+v^2= b^8+2b^5+b^2+2+b^4+2b^5+b^6

since uv=-1, and b^5=1

This gives b^4+b^3+b^2+b-6= 5

Hence  (u-v)^2= 5

u-v= sqrt(5)

The answer seems correct, but I'm wary of squaring as it normally adds complications to the matter... is there an easier and less time consuming way to do it?

Again thanks a lot for your help! With my math exam 2 weeks away, its a life-saver!

--- End quote ---

I can't think of a simpler way to do it.

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