Polynomial Question (1 Viewer)

gahyunkk · Apr 6, 2014

Can someone plz explain this to me? I dont know why i cant solve that equation
(

Especially bcuz of the -1 at the end)

Trebla · Apr 6, 2014

Let x = cos θ. The equation then becomes cos 5θ = 1. Solve for θ which allows you to solve for x. To deduce the exact values, use properties which link the roots and coefficients of polynomials.

Sy123 · Apr 6, 2014

gahyunkk · Apr 14, 2014

Sy123 said:
$\\ $After arriving at the equation$ \ \ \cos 5\theta = 16\cos^5 \theta - 20\cos^3 \theta + 5\cos \theta \\ $To arrive at the polynomial equation in$ \ x \ $it is clear that we must choose$ \ \theta \ $so that$ \ \cos 5\theta = 1 \\ \\ $Therefore the equation$ \ \ 16x^5-20x^3+5x - 1 = 0 \ $has roots which are the solutions to$ \ \ \cos 5\theta = 1 \ \ \ $The solutions are$ \ \ \ \cos \frac{2\pi}{5} , \cos \frac{4\pi}{5} , \cos \frac{6\pi}{5} , \cos \frac{8\pi}{5} , 1$

$\\ $However note that$ \ \cos \frac{6\pi}{5} = \cos \left(2\pi - \frac{4\pi}{5} \right) = \cos \frac{4\pi}{5} \ \ $and$ \ \cos \frac{8\pi}{5} = \cos \left(2\pi - \frac{2\pi}{5} \right) = \cos \frac{2\pi}{5} \\ \\ $Hence the roots of the equation in$ \ x \ $are a double root in$ \ \alpha = \cos \frac{2\pi}{5} \ $a double root in$ \ \beta = \cos \frac{4\pi}{5} \ $and$ \ 1 \\ \\ $Hence taking sum of roots$ \ \ \alpha + \beta + \alpha + \beta + 1 = 0 \Rightarrow \ \alpha + \beta = -\frac{1}{2} \ \ \ (1) \\ \\ $Taking product of roots$ \ \ \alpha^2 \beta^2 = \frac{1}{16} \ \ \ (2) \\ \\ $Hence solve$ \ (1) \ $and$ \ (2) \ $simultaneously to arrive at a quadratic which should output the desired result$$

Thank you so much sy123

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Polynomial Question (1 Viewer)

gahyunkk

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Trebla

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Sy123

This too shall pass

gahyunkk

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