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Question Number 108316 by bemath last updated on 16/Aug/20
((▽BeMath▽)/△) If x^4 +x^2 = ((11)/5) , find the value of Ω = (((x+1)/(x−1)))^(1/3) + (((x−1)/(x+1)))^(1/3)
$$\:\:\:\:\frac{\bigtriangledown\mathcal{B}{e}\mathcal{M}{ath}\bigtriangledown}{\bigtriangleup} \\ $$$${If}\:{x}^{\mathrm{4}} +{x}^{\mathrm{2}} \:=\:\frac{\mathrm{11}}{\mathrm{5}}\:,\:{find}\:{the}\:{value}\:{of} \\ $$$$\Omega\:=\:\sqrt[{\mathrm{3}}]{\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}}\:+\:\sqrt[{\mathrm{3}}]{\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}}\: \\ $$
Answered by bobhans last updated on 16/Aug/20
((bobhans)/𝛙) consider (x)^(1/3) +(y)^(1/3) = ((x+y)/( (x^2 )^(1/3) −((xy))^(1/3) +(y^2 )^(1/3) )) now (((x+1)/(x−1)))^(1/3) +(((x−1)/(x+1)))^(1/3) = ((((x+1)/(x−1))+((x−1)/(x+1)))/( (((((x+1)/(x−1)))^2 ))^(1/3) +(((((x−1)/(x+1)))^2 ))^(1/3) −1)) =((2x^2 +2)/((x^2 −1){(((((x+1)/(x−1)))^2 ))^(1/3) +(((((x−1)/(x+1)))^2 ))^(1/3) −1})) (∗)from the equation x^4 +x^2 =((11)/5) let x^2 =z ⇒5z^2 +5z−11=0 z= ((−5 ± (√(25+220)))/(10)) = ((−5 ± 7(√5))/(10)) if x∈R , we take z = ((−5 + 7(√5))/(10)) ≈ 1.065 and x = (√(1.065)) ≈ 1.032
$$\:\:\:\:\frac{\boldsymbol{{bobhans}}}{\boldsymbol{\psi}} \\ $$$${consider}\:\sqrt[{\mathrm{3}}]{{x}}\:+\sqrt[{\mathrm{3}}]{{y}}\:=\:\frac{{x}+{y}}{\:\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} }−\sqrt[{\mathrm{3}}]{{xy}}+\sqrt[{\mathrm{3}}]{{y}^{\mathrm{2}} }} \\ $$$${now}\:\sqrt[{\mathrm{3}}]{\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}}+\sqrt[{\mathrm{3}}]{\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}}\:=\:\frac{\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}+\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}}{\:\sqrt[{\mathrm{3}}]{\left(\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}\right)^{\mathrm{2}} }+\sqrt[{\mathrm{3}}]{\left(\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\right)^{\mathrm{2}} }−\mathrm{1}} \\ $$$$=\frac{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}}{\left({x}^{\mathrm{2}} −\mathrm{1}\right)\left\{\sqrt[{\mathrm{3}}]{\left(\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}\right)^{\mathrm{2}} }+\sqrt[{\mathrm{3}}]{\left(\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\right)^{\mathrm{2}} }−\mathrm{1}\right\}} \\ $$$$\left(\ast\right){from}\:{the}\:{equation}\:{x}^{\mathrm{4}} +{x}^{\mathrm{2}} =\frac{\mathrm{11}}{\mathrm{5}} \\ $$$${let}\:{x}^{\mathrm{2}} ={z}\:\Rightarrow\mathrm{5}{z}^{\mathrm{2}} +\mathrm{5}{z}−\mathrm{11}=\mathrm{0} \\ $$$${z}=\:\frac{−\mathrm{5}\:\pm\:\sqrt{\mathrm{25}+\mathrm{220}}}{\mathrm{10}}\:=\:\frac{−\mathrm{5}\:\pm\:\mathrm{7}\sqrt{\mathrm{5}}}{\mathrm{10}} \\ $$$${if}\:{x}\in\mathbb{R}\:,\:{we}\:{take}\:{z}\:=\:\frac{−\mathrm{5}\:+\:\mathrm{7}\sqrt{\mathrm{5}}}{\mathrm{10}}\:\approx\:\mathrm{1}.\mathrm{065} \\ $$$${and}\:{x}\:=\:\sqrt{\mathrm{1}.\mathrm{065}}\:\approx\:\mathrm{1}.\mathrm{032} \\ $$$$ \\ $$
Answered by Rasheed.Sindhi last updated on 16/Aug/20
x^4 +x^2 = ((11)/5)⇒x^2 =((−5±7(√5))/(10)) Ω = (((x+1)/(x−1)))^(1/3) + (((x−1)/(x+1)))^(1/3) Ω^3 =( (((x+1)/(x−1)))^(1/3) + (((x−1)/(x+1)))^(1/3) )^3 =((x+1)/(x−1))+((x−1)/(x+1))+3((((x+1)/(x−1)))^(1/3) + (((x−1)/(x+1)))^(1/3) ) Ω^3 −3Ω=((2(x^2 +1))/(x^2 −1)) =2(((((−5±7(√5))/(10))+1)/(((−5±7(√5))/(10))−1)))=2(((5±7(√5))/(−15±7(√5)))) =2(((5±7(√5))/(−15±7(√5)))×((−15∓7(√5))/(−15∓7(√5)))) =2(((−75∓35(√5)∓105(√5) −49(5))/(225−49(5)))) =((320±140(√5) )/(10))=32±14(√5)
$${x}^{\mathrm{4}} +{x}^{\mathrm{2}} \:=\:\frac{\mathrm{11}}{\mathrm{5}}\Rightarrow{x}^{\mathrm{2}} =\frac{−\mathrm{5}\pm\mathrm{7}\sqrt{\mathrm{5}}}{\mathrm{10}} \\ $$$$\Omega\:=\:\sqrt[{\mathrm{3}}]{\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}}\:+\:\sqrt[{\mathrm{3}}]{\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}}\: \\ $$$$\Omega^{\mathrm{3}} \:=\left(\:\sqrt[{\mathrm{3}}]{\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}}\:+\:\sqrt[{\mathrm{3}}]{\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}}\:\right)^{\mathrm{3}} \\ $$$$=\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}+\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}+\mathrm{3}\left(\sqrt[{\mathrm{3}}]{\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}}\:+\:\sqrt[{\mathrm{3}}]{\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}}\:\right) \\ $$$$\Omega^{\mathrm{3}} −\mathrm{3}\Omega=\frac{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{{x}^{\mathrm{2}} −\mathrm{1}}\:\:\:\:\:\:\:\:\: \\ $$$$\:\:=\mathrm{2}\left(\frac{\frac{−\mathrm{5}\pm\mathrm{7}\sqrt{\mathrm{5}}}{\mathrm{10}}+\mathrm{1}}{\frac{−\mathrm{5}\pm\mathrm{7}\sqrt{\mathrm{5}}}{\mathrm{10}}−\mathrm{1}}\right)=\mathrm{2}\left(\frac{\mathrm{5}\pm\mathrm{7}\sqrt{\mathrm{5}}}{−\mathrm{15}\pm\mathrm{7}\sqrt{\mathrm{5}}}\right) \\ $$$$\:\:\:=\mathrm{2}\left(\frac{\mathrm{5}\pm\mathrm{7}\sqrt{\mathrm{5}}}{−\mathrm{15}\pm\mathrm{7}\sqrt{\mathrm{5}}}×\frac{−\mathrm{15}\mp\mathrm{7}\sqrt{\mathrm{5}}}{−\mathrm{15}\mp\mathrm{7}\sqrt{\mathrm{5}}}\right) \\ $$$$\:\:=\mathrm{2}\left(\frac{−\mathrm{75}\mp\mathrm{35}\sqrt{\mathrm{5}}\mp\mathrm{105}\sqrt{\mathrm{5}}\:−\mathrm{49}\left(\mathrm{5}\right)}{\mathrm{225}−\mathrm{49}\left(\mathrm{5}\right)}\right) \\ $$$$=\frac{\mathrm{320}\pm\mathrm{140}\sqrt{\mathrm{5}}\:}{\mathrm{10}}=\mathrm{32}\pm\mathrm{14}\sqrt{\mathrm{5}} \\ $$
Commented by bemath last updated on 16/Aug/20
coll and nice
$${coll}\:{and}\:{nice} \\ $$
Answered by mr W last updated on 16/Aug/20
x^4 +x^2 −((11)/5)=0 x^2 =(1/2)(−1+(√(1+((44)/5))))=((7(√5)−5)/(10))>0 ((x^2 +1)/(x^2 −1))=((7(√5)+5)/(7(√5)−15))=16+7(√5) (a+b)^3 =a^3 +b^3 +3ab(a+b) let a=(((x+1)/(x−1)))^(1/3) , b=(((x−1)/(x+1)))^(1/3) Ω^3 =2×((x^2 +1)/(x^2 −1))+3Ω ⇒Ω^3 −3Ω−2λ=0 with λ=((x^2 +1)/(x^2 −1))=16+7(√5) Δ=(−1)^3 +(−λ)^2 =4(125+56(√5))>0 Ω=(((√Δ)+λ))^(1/3) −(((√Δ)−λ))^(1/3) =((2(√(125+56(√5)))+16+7(√5)))^(1/3) −((2(√(125+56(√7)))−16−7(√5)))^(1/3) ≈4.236068
$${x}^{\mathrm{4}} +{x}^{\mathrm{2}} −\frac{\mathrm{11}}{\mathrm{5}}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\left(−\mathrm{1}+\sqrt{\mathrm{1}+\frac{\mathrm{44}}{\mathrm{5}}}\right)=\frac{\mathrm{7}\sqrt{\mathrm{5}}−\mathrm{5}}{\mathrm{10}}>\mathrm{0} \\ $$$$\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{1}}=\frac{\mathrm{7}\sqrt{\mathrm{5}}+\mathrm{5}}{\mathrm{7}\sqrt{\mathrm{5}}−\mathrm{15}}=\mathrm{16}+\mathrm{7}\sqrt{\mathrm{5}} \\ $$$$ \\ $$$$\left({a}+{b}\right)^{\mathrm{3}} ={a}^{\mathrm{3}} +{b}^{\mathrm{3}} +\mathrm{3}{ab}\left({a}+{b}\right) \\ $$$${let}\:{a}=\sqrt[{\mathrm{3}}]{\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}},\:{b}=\sqrt[{\mathrm{3}}]{\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}} \\ $$$$\Omega^{\mathrm{3}} =\mathrm{2}×\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{1}}+\mathrm{3}\Omega \\ $$$$\Rightarrow\Omega^{\mathrm{3}} −\mathrm{3}\Omega−\mathrm{2}\lambda=\mathrm{0}\:{with}\:\lambda=\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{1}}=\mathrm{16}+\mathrm{7}\sqrt{\mathrm{5}} \\ $$$$\Delta=\left(−\mathrm{1}\right)^{\mathrm{3}} +\left(−\lambda\right)^{\mathrm{2}} =\mathrm{4}\left(\mathrm{125}+\mathrm{56}\sqrt{\mathrm{5}}\right)>\mathrm{0} \\ $$$$\Omega=\sqrt[{\mathrm{3}}]{\sqrt{\Delta}+\lambda}−\sqrt[{\mathrm{3}}]{\sqrt{\Delta}−\lambda} \\ $$$$=\sqrt[{\mathrm{3}}]{\mathrm{2}\sqrt{\mathrm{125}+\mathrm{56}\sqrt{\mathrm{5}}}+\mathrm{16}+\mathrm{7}\sqrt{\mathrm{5}}}−\sqrt[{\mathrm{3}}]{\mathrm{2}\sqrt{\mathrm{125}+\mathrm{56}\sqrt{\mathrm{7}}}−\mathrm{16}−\mathrm{7}\sqrt{\mathrm{5}}} \\ $$$$\approx\mathrm{4}.\mathrm{236068} \\ $$

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