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a) P(iw)=4w^8-17w^4+4=P(w) As w is a root of P(z) ie P(w)=0. It follows P(iw)=0 P(1/w)=4(1/w^8)-17/w^4+4 =[4w^8-17w^4+4]/w^8 =P(w)/w^8 =0 b) Let y=w^4: P(y)=4y^2-17y+4 y=[+17+-sqrt(17^2-64)]/8 y=4or 1/4 w=+-1/sqrt(2), +-sqrt(2), +-isqrt(2), +-i/sqrt(2)

Cancel every odd term: (z-1)^6+(z+1)^6=0 2(z^6+15z^4+15z^2+1)=0 2(z^2+1)(z^4+14z^2+1)=0 Thus, z=+-i OR z^2=(-14+-sqrt(14^2-4))/2 z^2=-(7+-4sqrt(3)) Thus, z=+-isqrt(7+-4sqrt(3))

I don't know, as I haven't worked on the problem yet (and don't plan to). I was just suggesting another method to calculate the areas as a iterative process wouldn't have to be used in the method. @ would just be atan(y/x) where y,x are co-ords of C, i think.

If you have the Coordinates of C and where the larger circle meets the x-axis near B. Could you just use A=1/2 r^2(@-sin@) in both segments by constructing BD,BC,AD,AC?

Shit, I always miss out on integration Q's

b^2=ac a^2+ac=c^2 a^2+ac-c^2=0 (a+[c/2])^2=5c^2/4 a+[c/2]=sqrt(5)c/2 a=(c/2)[sqrt(5)-1] a[sqrt(5)+1]/2=c c/a=[sqrt(5)+1]/2 Idk, if this is correct though

p-1=[(x+a)/2a]-1 p-1=(x+a-2a)/2a p-1=(x-a)/2a

P(2ap,ap^2) and Q(2ap-2a,aq^2) ie. 2aq=2ap-2a => q=p-1 => q=(x-a)/2a Midpt: (2ap-a,a/2(p^2+q^2)) x=2ap-a x+a=2ap p=(x+a)/2a ...sub into y equ y=a/2(p^2+q^2) y=(a/2)((x+a)^2/4a^2+((x-a)/4a^2)) y=(1/8a)((x+a)^2+(x-a)^2) y=(1/4a)(x^2+a^2) x^2=4ay-a^2

Yeah, i usually do this but untouchablecuz's method is also good

Oh, lol.

Yeah maybe, but it takes a while to find x=-3/2 and x=4/3 as roots as there are many possible pairings. But, idk it could work if you have that much time.

Yeah, nw. Your username looks cool now :)

My working follows the same. After this i tried factor theorem using rational roots theorem to find my pairs. I then subbed in x=-1.5 into either of the summation pairs for a+b or c+d. Then just factor the poly with the solutions you get for c and d.

Re: Quartic Roots Let roots be; a,b,c,d. ab=cd=-2. a+b= 2 c+d= -1/6 After trying factor theorem more than 10 times. I get x=-1.5 or -3/2 as a root. Checking solutions on my calculator: x=3.5 isn't a root when subbing in for a or b. Thus c or d=4/3 is another root Now factoring the poly...

Yeah, a lot less tedious solution but that was a really good solution.

Conrats on becoming a Mod

did you know the answers for y and z before you wrote your solution?

(a^2+b^2/2ab)^2-(a^2-b^2/2ab)^2=(b/a)(a/b)=1

Nice work Gurmies :)

1) The easiest way is to solve the longer quadratic equation using the quadratic formula. 2) Using the fact that a+b=-m and ab=n, sub these into x=[m^2-2n+-msqrt(m^2-4n)]/2n which you attained in 1) It works, trust me. To use z+y=[a^2+b^2]/ab would be taking the longer solution ie don't use...