If-x-1-17-2-Find-the-value-of-x-3-2x-2-7x-1-x-2-x-1-decimal-point-

Question Number 17270 by VEGAMIND last updated on 03/Jul/17

If x = \frac{1 + \sqrt{17}}{2} . Find the value of

\frac{x^{3} - 2 x^{2} + 7 x - 1}{x^{2} - x + 1} decimal point .

Commented by 18±1 last updated on 03/Jul/17

. 2 x = 1 + \sqrt{17}

2 x - 1 = \sqrt{17}

4 x^{2} - 4 x + 1 = 17

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

. x^{3} - 2 x^{2} + 7 x - 1 = x^{2} - 5 x - 2 x^{2} + 7 x - 1

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

= - {(x - 1)}^{2}

2 x - 1 = \sqrt{17}

2 x - 2 = \sqrt{17} - 1

\Rightarrow x - 1 = \frac{\sqrt{17} - 1}{2}

\Rightarrow \frac{x^{3} - 2 x^{2} + 7 x - 1}{x^{2} - x + 1} = \frac{5}{- {(\frac{\sqrt{17} - 1}{2})}^{2}}

Commented by 18±1 last updated on 03/Jul/17

s o r r y I w r o n g a t x^{2} - x + 1 = 5

\Rightarrow x^{3} = x^{2} - 5 x

r i g h t i s x^{3} = x^{2} + 4 x

u s e t h i s t e c n i p, l e t y o u t r y a g a i n

Answered by mrW1 last updated on 03/Jul/17

x = \frac{1 + \sqrt{17}}{2}

2 x - 1 = \sqrt{17}

{4 x}^{2} - 4 x + 1 = 17

{4 x}^{2} - 4 x + 4 = 20

x^{2} - x + 1 = 5

x^{2} - x = 4

x^{3} - x^{2} = 4 x

x^{3} - {2 x}^{2} + 7 x - 1 = 4 x - x^{2} + 7 x - 1

x^{3} - {2 x}^{2} + 7 x - 1 = - x^{2} + 11 x - 1

x^{3} - {2 x}^{2} + 7 x - 1 = - (x^{2} - x + 1) + 10 x

x^{3} - {2 x}^{2} + 7 x - 1 = - 5 + 10 x

x^{3} - {2 x}^{2} + 7 x - 1 = 5 (2 x - 1)

x^{3} - {2 x}^{2} + 7 x - 1 = 5 \sqrt{17}

\Rightarrow \frac{x^{3} - 2 x^{2} + 7 x - 1}{x^{2} - x + 1} = \frac{5 \sqrt{17}}{5} = \sqrt{17} \approx 4.123

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