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- Feb 14, 2012

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- Thread starter anemone
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- Feb 14, 2012

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- Mar 31, 2013

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In the link provided by opalgs above

the roots of the equation t^3−3√3t^2−3t+√3=0

will be tan20,tan(−40)=−tan40,tan80

that is of f(t) =t^3−3√3t^2−3t+√3=0

and we shall construct an equation whose roots are

tan^2 20,tan^2 40 tan^2 80

shall be f(x^(1/2) = 0)

putting t = x^(1/2) we get

so x^(3/2) - 3√3x−3x^(1/2) +√3=0

or x^(3/2) - 3 x^(1/2) = 3√3x -√3

or

√x(x-3) = √3(3x -1)

square both sides to get

x(x-3)^3 = 3(3x-1)^2

or x(x^2-6x+ 9) = 3(9x^2 - 6x + 1)

or x^3 - 33 x^2 + 27x - 3 = 0

as it is cubic roots are tan^2 20,tan^2 40 tan^2 80

and sum of roots = - coefficent of x^2 or 33