If-x-7-4-3-Find-x-4-x-3-9x-2-2x-5-x-2-4x-3-

Question Number 199550 by hardmath last updated on 05/Nov/23

If

x = \sqrt{7 + 4 \sqrt{3}}

Find :

\frac{x^{4} - x^{3} - {9 x}^{2} - 2 x + 5}{x^{2} - 4 x + 3} = ?

Answered by Rasheed.Sindhi last updated on 07/Nov/23

x = \sqrt{7 + 4 \sqrt{3}}; \frac{x^{4} - x^{3} - {9 x}^{2} - 5 x + 5}{x^{2} - 4 x + 3} = ?

x = \sqrt{7 + 4 \sqrt{3}} = 2 + \sqrt{3}^{★}

∙ x = 2 + \sqrt{3}

{(x - 2)}^{2} = 3

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

∙ x^{2} = 4 x - 1

∙ x^{3} = {4 x}^{2} - x = 4 (4 x - 1) - x = 15 x - 4

∙ x^{4} = {15 x}^{2} - 4 x = 15 (4 x - 1) - 4 x = 56 x - 15

\frac{x^{4} - x^{3} - {9 x}^{2} - 5 x + 5}{x^{2} - 4 x + 3}

= \frac{(56 x - 15) - (15 x - 4) - 9 (4 x - 1) - 5 x + 5}{(4 x - 1) - 4 x + 3}

= \frac{56 x - 15 - 15 x + 4 - 36 x + 9 - 5 x + 5}{4 x - 1 - 4 x + 3}

= \frac{41 x - 15 + 4 - 41 x + 9 - 5 x + 5}{4 x - 1 - 4 x + 3} = \frac{- 11 + 14}{2} = \frac{3}{2}

Commented by Rasheed.Sindhi last updated on 06/Nov/23

^{★}

x = \sqrt{7 + 4 \sqrt{3}} = a + b \sqrt{3}

a^{2} + {3 b}^{2} + 2 ab \sqrt{3} = 7 + 4 \sqrt{3}

a^{2} + {3 b}^{2} = 7 \land ab = 2 \Rightarrow b = 2 / a

a^{2} + 3 {(\frac{2}{a})}^{2} = 7

a^{4} - {7 a}^{2} + 12

a^{2} = \frac{7 \pm \sqrt{49 - 48}}{2} = 4, \overset{\times}{3} \Rightarrow a = \pm 2 \Rightarrow b = \pm 1

x = \sqrt{7 + 4 \sqrt{3}} = 2 + \sqrt{3}, - 2 - \sqrt{3} (R e j e c t e d)

Answered by AST last updated on 05/Nov/23

\frac{x^{4} - x^{3} - 9 x^{2} - 2 x + 5 = x^{2} (x^{2} - 4 x + 3) + 3 x (x^{2} - 4 x + 3) + 5 - 14 x}{x^{2} - 4 x + 3}

= x^{2} + 3 x + \frac{5 - 14 x}{(x - 1) (x - 3)}

= 7 + 4 \sqrt{3} + 3 (2 + \sqrt{3}) + \frac{- 23 - 14 \sqrt{3}}{2} = 1.5 = \frac{3}{2}

Commented by hardmath last updated on 05/Nov/23

Sorry professor :

\frac{x^{4} - x^{3} - {9 x}^{2} - 5 x + 5}{x^{2} - 4 x + 3}

Commented by hardmath last updated on 05/Nov/23

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