On the roots of t^4-6\sqrt{3}t^3+8t^2+2\sqrt{3}t-1=0

Elise Winters

Elise Winters

Answered question

2022-04-22

On the roots of
t463t3+8t2+23t1=0

Answer & Explanation

ubafumene42h

ubafumene42h

Beginner2022-04-23Added 13 answers

First of all, tan5θ=5tanθ103θ+tan5θ110tan2θ+5tan4θ
Set tan5θ=3=tanπ3
5θ=nπ+π3=(3n+1)π5 where n is any integer
So, the values of θ can be set from n=0,1,2,3,4
Observe that for n=3,tanθ=tan2π3=tan(ππ3)=tanπ3=3
So rearrange 3=5tanθ10tan3θ+tan5θ110tan2θ+5tan4θ and divide by
tanθ(3)=tanθ+3 to get the equation with roots tan(3n+1)π15 where n=0,1,2,4

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