Question-47986

Question Number 47986 by behi83417@gmail.com last updated on 17/Nov/18

Commented by maxmathsup by imad last updated on 18/Nov/18

2) (e) \Leftrightarrow \frac{1 - x^{8}}{1 - x} = 0 \Leftrightarrow x^{8} = 1 a n d x \neq 1 x = r e^{i θ} \Rightarrow r^{8} e^{i 8 θ} = e^{i 2 k π} \Rightarrow

r = 1 a n d 8 θ = 2 k π \Rightarrow x_{k} = e^{i \frac{k π}{4}} w i t h k \in [[1, 7]] a r e t h e r o o t s o f t h i s e q u a t i o n .

Answered by tanmay.chaudhury50@gmail.com last updated on 18/Nov/18

1) x^{4} + x^{3} + x^{2} + x + 1 = 0

x^{2} + x + 1 + \frac{1}{x} + \frac{1}{x^{2}} = 0

(x^{2} + \frac{1}{x^{2}}) + (x + \frac{1}{x}) + 1 = 0

y = x + \frac{1}{x}

y^{2} - 2 + y + 1 = 0

y^{2} + y - 1 = 0

y = \frac{- 1 \pm \sqrt{1 + 4}}{2} = \frac{- 1 \pm \sqrt{5}}{2} = \frac{- 1 + \sqrt{5}}{2} a n d \frac{- 1 - \sqrt{5}}{2}

s o x + \frac{1}{x} = y_{1} a n d x + \frac{1}{x} = y_{2}

y_{1} = \frac{- 1 + \sqrt{5}}{2} a n d y_{2} = \frac{- 1 - \sqrt{5}}{2}

x + \frac{1}{x} = y_{1}

x^{2} + 1 = {x y}_{1}

x^{2} - {x y}_{1} + 1 = 0

x = \frac{y_{1} \pm \sqrt{y_{1}^{2} - 4}}{2} = \frac{\frac{- 1 + \sqrt{5}}{2} \pm \sqrt{\frac{1 - 2 \sqrt{5 + 5}}{4} - 4}}{2}

x = \frac{\frac{- 1 + \sqrt{5}}{2} \pm \sqrt{\frac{1 - 2 \sqrt{5 + 5 - 16}}{4}}}{2}

x = \frac{\frac{- 1 + \sqrt{5}}{2} \pm \sqrt{\frac{- 10 - 2 \sqrt{5}}{4}}}{2}

x = \frac{- 1 + \sqrt{5} \pm i \sqrt{10 + 2 \sqrt{5}}}{4}

w h e n x + \frac{1}{x} = y_{2}

x^{2} - {x y}_{2} + 1 = 0

x = \frac{y_{2} \pm \sqrt{y_{2}^{2} - 4}}{2}

x = \frac{\frac{- 1 - \sqrt{5}}{2} \pm \sqrt{\frac{1 + 2 \sqrt{5 + 5}}{4} - 4}}{2}

= \frac{\frac{- 1 - \sqrt{5}}{2} \pm \sqrt{\frac{- 10 + 2 \sqrt{5}}{4}}}{2}

= \frac{- 1 - \sqrt{5} \pm i \sqrt{10 - 2 \sqrt{5}}}{4}

Answered by tanmay.chaudhury50@gmail.com last updated on 18/Nov/18

o r s o l u t i o n

x^{4} + x^{3} + x^{2} + x + 1 = 0

(x - 1) (x^{4} + x^{3} + x^{2} + x + 1) = 0

x^{5} - 1 = 0

x^{5} = 1 = c o s 2 k π + i s i n 2 π

x = c o s (\frac{2 k π}{5}) + i s i n (\frac{2 k π}{5})

s o r o o t s a r e \dots p u t k = 0, \pm 1, \pm 2

Answered by tanmay.chaudhury50@gmail.com last updated on 18/Nov/18

2) x^{6} (x + 1) + x^{4} (x + 1) + x^{2} (x + 1) + 1 (x + 1) = 0

(x + 1) (x^{6} + x^{4} + x^{2} + 1) = 0

(x + 1) {x^{4} (x^{2} + 1) + 1 (x^{2} + 1)} = 0

(x + 1) (x^{2} + 1) (x^{4} + 1) = 0

w h e n x + 1 = 0

x = - 1

x^{2} + 1 = 0

x^{2} = - 1 = i^{2}

x = \pm i

x^{4} + 1 = 0

x^{4} = - 1 = c o s (2 k π + π) + i s i n (2 k π + π)

x = c o s (\frac{2 k + 1}{4} π) + i s i n (\frac{2 k + 1}{4} π)