Given-x-1-x-5-then-x-4-1-x-4-x-2-3x-1-

Question Number 135418 by liberty last updated on 13/Mar/21

G i v e n x + \frac{1}{x} = 5 t h e n \frac{x^{4} + \frac{1}{x^{4}}}{x^{2} - 3 x + 1} ?

Answered by SEKRET last updated on 13/Mar/21

{(x + \frac{1}{x})}^{2} = 5^{2} x^{2} + \frac{1}{x^{2}} = 23 {(x^{2} + \frac{1}{x^{2}})}^{2} = 23^{2}

x^{4} + \frac{1}{x^{4}} = 23^{2} - 2 = 527

x^{2} - 3 x + 1 = ? x^{2} - 5 x + 1 = 0

x^{2} + 1 = 5 x x^{2} + 1 - 3 x = 2 x

\frac{x^{4} + \frac{1}{x^{4}}}{x^{2} - 3 x + 1} = \frac{527}{2 x} x^{2} - 5 x + 1 = 0 x = \frac{5 \mp \sqrt{21}}{2}

\frac{x^{4} + \frac{1}{x^{4}}}{x^{2} - 3 x + 1} = \frac{527}{5 \mp \sqrt{21}} = \frac{2635 \mp 527 \sqrt{21}}{4}

Answered by Ñï= last updated on 13/Mar/21

x + \frac{1}{x} = 5

\Rightarrow x^{2} + \frac{1}{x^{2}} = 23 \Rightarrow x^{4} + \frac{1}{x^{4}} = 23^{2} - 2 = 527

x^{2} - 5 x + 1 = 0

\Rightarrow x = \frac{5 \pm \sqrt{21}}{2} \Rightarrow 2 x = 5 \pm \sqrt{21}

\frac{\frac{1}{x} (x^{4} + \frac{1}{x^{4}})}{x + \frac{1}{x} - 3} = \frac{527}{2 x} = \frac{527}{5 \pm \sqrt{21}} = \frac{527}{4} (5 \pm \sqrt{21})

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