Question-145114

Question Number 145114 by imjagoll last updated on 02/Jul/21

Commented by MJS_new last updated on 02/Jul/21

for x \in R I get 2 + \sqrt{5}

Commented by imjagoll last updated on 02/Jul/21

how did the answer sir ?

Commented by Canebulok last updated on 02/Jul/21

R a n d o m P r o b l e m :

I f x^{4} + x^{2} = \frac{11}{5}

W h a t i s t h e v a l u e o f

^{3} \sqrt{\frac{x + 1}{x - 1}} +^{3} \sqrt{\frac{x - 1}{x + 1}} = ?

S o l u t i o n :

\Rightarrow 5 x^{4} + 5 x^{2} - 11 = 0

l e t :

\Rightarrow x^{2} = K

∵

\Rightarrow 5 K^{2} + 5 K - 11 = 0

B y q u a d r a t i c f o r m u l a,

\Rightarrow K = \frac{- 5 \pm \sqrt{5^{2} - 4 (5) (- 11)}}{2 (5)}

\Rightarrow K_{1} = \frac{- 5 + 7 \sqrt{5}}{10}

\Rightarrow K_{2} = \frac{- 5 - 7 \sqrt{5}}{10}

∴

\Rightarrow x^{2} = \frac{- 5 \pm 7 \sqrt{5}}{10}

\Rightarrow x = \sqrt{(\frac{- 5 \pm 7 \sqrt{5}}{10})}

T h u s;

\Rightarrow^{3} \sqrt{\frac{x + 1}{x - 1}} +^{3} \sqrt{\frac{x - 1}{x + 1}} \approx 1.837977

o r

\Rightarrow^{3} \sqrt{\frac{x + 1}{x - 1}} +^{3} \sqrt{\frac{x - 1}{x + 1}} \approx 4.236068

Commented by Canebulok last updated on 02/Jul/21

sorry i didn't notice

Answered by liberty last updated on 02/Jul/21

\Rightarrow 5 x^{4} + 5 x^{2} - 11 = 0

\Rightarrow x^{2} = \frac{- 5 + \sqrt{25 + 220}}{10}

\Rightarrow x^{2} + 1 = \frac{5 + \sqrt{245}}{10}

\Rightarrow x^{2} + 1 = \frac{5 + 7 \sqrt{5}}{10} \dots (i)

\Rightarrow x^{2} - 1 = \frac{- 15 + 7 \sqrt{5}}{10} \dots (i i)

\Rightarrow \frac{x^{2} + 1}{x^{2} - 1} = \frac{5 + 7 \sqrt{5}}{- 15 + 7 \sqrt{5}} \dots (i i i)

c o n s i d e r \sqrt[3]{\frac{x + 1}{x - 1}} + \sqrt[3]{\frac{x - 1}{x + 1}}

= \frac{\frac{x + 1}{x - 1} + \frac{x - 1}{x + 1}}{\sqrt[3]{{(\frac{x + 1}{x - 1})}^{2}} + \sqrt[3]{{(\frac{x - 1}{x + 1})}^{2}} - 1}

= \frac{\frac{2 (x^{2} + 1)}{x^{2} - 1}}{\sqrt[3]{\frac{{(x + 1)}^{3}}{{(x^{2} - 1)}^{2}}} + \sqrt[3]{\frac{{(x - 1)}^{3}}{{(x^{2} - 1)}^{2}}} - 1}

= \frac{2 (\frac{7 \sqrt{5} + 5}{7 \sqrt{5} - 15})}{\frac{2 x}{\sqrt[3]{{(\frac{7 \sqrt{5} - 15}{10})}^{2}}} - 1}

\dots

Answered by MJS_new last updated on 02/Jul/21

a^{1 / 3} + b^{1 / 3} = c

a + 3 a^{1 / 3} b^{1 / 3} (a^{1 / 3} + b^{1 / 3}) + b = c^{3}

a + 3 a^{1 / 3} b^{1 / 3} c + b = c^{3}

3 a^{1 / 3} b^{1 / 3} c = c^{3} - a - b

27 {a b c}^{3} = {(c^{3} - a - b)}^{3}

c^{9} - 3 (a + b) c^{6} + 3 (a^{2} - 7 a b + b^{2}) c^{3} - {(a + b)}^{3} = 0

a = \frac{x + 1}{x - 1} \land b = \frac{1}{a}

c^{9} - \frac{6 (x^{2} + 1)}{x^{2} - 1} c^{6} - \frac{3 (x^{2} - 5) (5 x^{2} - 1)}{{(x^{2} - 1)}^{2}} c^{3} - \frac{8 {(x^{2} + 1)}^{3}}{{(x^{2} - 1)}^{3}} = 0

x^{4} + x^{2} = \frac{11}{5} \Rightarrow x^{2} = - \frac{1}{2} + \frac{7 \sqrt{5}}{10}

c^{9} - 6 (16 + 7 \sqrt{5}) c^{6} + 21 (285 + 128 \sqrt{5}) c^{3} - 8 (15856 + 7091 \sqrt{5}) = 0

c = {(z + 2 (16 + 7 \sqrt{5}))}^{1 / 3}

z^{3} - 27 z - 54 (16 + 7 \sqrt{5}) = 0

(z - 3 (2 + \sqrt{5})) (z^{2} + 3 (2 + \sqrt{5}) z + 18 (3 + 2 \sqrt{5})) = 0

\Rightarrow

z = 3 (2 + \sqrt{5}) \Rightarrow c = \sqrt[3]{38 + 17 \sqrt{5}} = 2 + \sqrt{5}

Commented by MJS_new last updated on 02/Jul/21

is this really so hard ? I {can}^{'} t believe \dots

Commented by imjagoll last updated on 03/Jul/21

hahaha i believe it problem super

hard sir