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Question Number 137648 by JulioCesar last updated on 05/Apr/21

Commented by mr W last updated on 05/Apr/21

2+(√5)

$$\mathrm{2}+\sqrt{\mathrm{5}} \\ $$

Answered by mr W last updated on 05/Apr/21

x^4 +x^2 −((11)/5)=0 ⇒x^2 =((7(√5)−5)/( 10)) p=(((x+1)/(x−1)))^(1/3) +(((x−1)/(x+1)))^(1/3) p^3 =((x+1)/(x−1))+((x−1)/(x+1))+3p p^3 =2(1+(2/(x^2 −1)))+3p p^3 =2(1+(2/(((7(√5)−5)/( 10))−1)))+3p p^3 =2(7(√5)+16)+3p p^3 −3p−2(7(√5)+16)=0 p=2+(√5) (real root)

$${x}^{\mathrm{4}} +{x}^{\mathrm{2}} −\frac{\mathrm{11}}{\mathrm{5}}=\mathrm{0} \\ $$$$\Rightarrow{x}^{\mathrm{2}} =\frac{\mathrm{7}\sqrt{\mathrm{5}}−\mathrm{5}}{\:\mathrm{10}} \\ $$$${p}=\sqrt[{\mathrm{3}}]{\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}}+\sqrt[{\mathrm{3}}]{\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}} \\ $$$${p}^{\mathrm{3}} =\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}+\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}+\mathrm{3}{p} \\ $$$${p}^{\mathrm{3}} =\mathrm{2}\left(\mathrm{1}+\frac{\mathrm{2}}{{x}^{\mathrm{2}} −\mathrm{1}}\right)+\mathrm{3}{p} \\ $$$${p}^{\mathrm{3}} =\mathrm{2}\left(\mathrm{1}+\frac{\mathrm{2}}{\frac{\mathrm{7}\sqrt{\mathrm{5}}−\mathrm{5}}{\:\mathrm{10}}−\mathrm{1}}\right)+\mathrm{3}{p} \\ $$$${p}^{\mathrm{3}} =\mathrm{2}\left(\mathrm{7}\sqrt{\mathrm{5}}+\mathrm{16}\right)+\mathrm{3}{p} \\ $$$${p}^{\mathrm{3}} −\mathrm{3}{p}−\mathrm{2}\left(\mathrm{7}\sqrt{\mathrm{5}}+\mathrm{16}\right)=\mathrm{0} \\ $$$${p}=\mathrm{2}+\sqrt{\mathrm{5}}\:\left({real}\:{root}\right) \\ $$

Commented by JulioCesar last updated on 05/Apr/21

thank sir

$${thank}\:{sir} \\ $$

Commented by otchereabdullai@gmail.com last updated on 05/Apr/21

nice one!

$$\mathrm{nice}\:\mathrm{one}! \\ $$

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