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Question Number 201679 by cherokeesay last updated on 10/Dec/23
Answered by Rasheed.Sindhi last updated on 10/Dec/23
((1−(√(x^2 +1))))^(1/6) +(((√(x^2 +1)) −1))^(1/6) =1 a+b=1⇒b=1−a a^6 +b^6 =1−(√(x^2 +1)) + (√(x^2 +1)) −1=0 (a^2 +b^2 )(a^4 −a^2 b^2 +b^4 )=0 a^2 +b^2 =0 ∣ a^4 −a^2 b^2 +b^4 =0^★ (a+b)^2 −2ab=0 1−2ab=0 1−2a(1−a)=0 2a^2 −2a+1=0 a=((2±(√(4−8)))/4)=((1±i)/2) ((1−(√(x^2 +1))))^(1/6) =((1±i)/2) 1−(√(x^2 +1)) =(((1±i)/2))^6 (√(x^2 +1)) =1−(((1±i)/2))^6 x^2 +1 =(1−(((1±i)/2))^6 )^2 x=±(√((1−(((1±i)/2))^6 )^2 −1)) =±(√(1−2(((1±i)/2))^6 +(((1±i)/2))^(12) −1)) =±(√(−2(((1±i)/2))^6 +(((1±i)/2))^(12) )) =±(((1±i)/2))^6 (√(−2+(((1±i)/2))^6 )) .... ...
$$\sqrt[{\mathrm{6}}]{\mathrm{1}−\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}\:+\sqrt[{\mathrm{6}}]{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:−\mathrm{1}}\:=\mathrm{1} \\ $$$${a}+{b}=\mathrm{1}\Rightarrow{b}=\mathrm{1}−{a} \\ $$$${a}^{\mathrm{6}} +{b}^{\mathrm{6}} =\mathrm{1}−\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:+\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:−\mathrm{1}=\mathrm{0} \\ $$$$\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left({a}^{\mathrm{4}} −{a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{4}} \right)=\mathrm{0} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{0}\:\mid\:{a}^{\mathrm{4}} −{a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{4}} =\mathrm{0}^{\bigstar} \\ $$$$\left({a}+{b}\right)^{\mathrm{2}} −\mathrm{2}{ab}=\mathrm{0} \\ $$$$\mathrm{1}−\mathrm{2}{ab}=\mathrm{0} \\ $$$$\mathrm{1}−\mathrm{2}{a}\left(\mathrm{1}−{a}\right)=\mathrm{0} \\ $$$$\mathrm{2}{a}^{\mathrm{2}} −\mathrm{2}{a}+\mathrm{1}=\mathrm{0} \\ $$$${a}=\frac{\mathrm{2}\pm\sqrt{\mathrm{4}−\mathrm{8}}}{\mathrm{4}}=\frac{\mathrm{1}\pm{i}}{\mathrm{2}} \\ $$$$\sqrt[{\mathrm{6}}]{\mathrm{1}−\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}\:=\frac{\mathrm{1}\pm{i}}{\mathrm{2}} \\ $$$$\mathrm{1}−\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:=\left(\frac{\mathrm{1}\pm{i}}{\mathrm{2}}\right)^{\mathrm{6}} \\ $$$$\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:=\mathrm{1}−\left(\frac{\mathrm{1}\pm{i}}{\mathrm{2}}\right)^{\mathrm{6}} \\ $$$${x}^{\mathrm{2}} +\mathrm{1}\:=\left(\mathrm{1}−\left(\frac{\mathrm{1}\pm{i}}{\mathrm{2}}\right)^{\mathrm{6}} \right)^{\mathrm{2}} \\ $$$${x}=\pm\sqrt{\left(\mathrm{1}−\left(\frac{\mathrm{1}\pm{i}}{\mathrm{2}}\right)^{\mathrm{6}} \right)^{\mathrm{2}} −\mathrm{1}}\: \\ $$$$\:\:\:=\pm\sqrt{\mathrm{1}−\mathrm{2}\left(\frac{\mathrm{1}\pm{i}}{\mathrm{2}}\right)^{\mathrm{6}} +\left(\frac{\mathrm{1}\pm{i}}{\mathrm{2}}\right)^{\mathrm{12}} −\mathrm{1}}\: \\ $$$$\:\:\:=\pm\sqrt{−\mathrm{2}\left(\frac{\mathrm{1}\pm{i}}{\mathrm{2}}\right)^{\mathrm{6}} +\left(\frac{\mathrm{1}\pm{i}}{\mathrm{2}}\right)^{\mathrm{12}} }\: \\ $$$$\:\:\:=\pm\left(\frac{\mathrm{1}\pm{i}}{\mathrm{2}}\right)^{\mathrm{6}} \sqrt{−\mathrm{2}+\left(\frac{\mathrm{1}\pm{i}}{\mathrm{2}}\right)^{\mathrm{6}} }\: \\ $$$$…. \\ $$$$… \\ $$
Answered by Rasheed.Sindhi last updated on 10/Dec/23
1−(√(x^2 +1)) =a (say) (a)^(1/6) +((−a))^(1/6) =1 ((a)^(1/6) +((−a))^(1/6) )^2 =1^2 a^(1/3) +2a^(1/6) (−a)^(1/6) +(−a)^(1/3) =1 a^(1/3) +(−a)^(1/3) =1−2a^(1/6) (−a)^(1/6) (a^(1/3) +(−a)^(1/3) )^3 =(1−2a^(1/6) (−a)^(1/6) )^3 {a+(−a)}+3(a^(1/3) )(−a)^(1/3) {a^(1/3) +(−a)^(1/3) } =.....
$$\mathrm{1}−\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:={a}\:\left({say}\right) \\ $$$$\sqrt[{\mathrm{6}}]{{a}}\:+\sqrt[{\mathrm{6}}]{−{a}}\:=\mathrm{1} \\ $$$$\left(\sqrt[{\mathrm{6}}]{{a}}\:+\sqrt[{\mathrm{6}}]{−{a}}\:\right)^{\mathrm{2}} =\mathrm{1}^{\mathrm{2}} \\ $$$${a}^{\mathrm{1}/\mathrm{3}} +\mathrm{2}{a}^{\mathrm{1}/\mathrm{6}} \left(−{a}\right)^{\mathrm{1}/\mathrm{6}} +\left(−{a}\right)^{\mathrm{1}/\mathrm{3}} =\mathrm{1} \\ $$$${a}^{\mathrm{1}/\mathrm{3}} +\left(−{a}\right)^{\mathrm{1}/\mathrm{3}} =\mathrm{1}−\mathrm{2}{a}^{\mathrm{1}/\mathrm{6}} \left(−{a}\right)^{\mathrm{1}/\mathrm{6}} \\ $$$$\left({a}^{\mathrm{1}/\mathrm{3}} +\left(−{a}\right)^{\mathrm{1}/\mathrm{3}} \right)^{\mathrm{3}} =\left(\mathrm{1}−\mathrm{2}{a}^{\mathrm{1}/\mathrm{6}} \left(−{a}\right)^{\mathrm{1}/\mathrm{6}} \right)^{\mathrm{3}} \\ $$$$\left\{{a}+\left(−{a}\right)\right\}+\mathrm{3}\left({a}^{\mathrm{1}/\mathrm{3}} \right)\left(−{a}\right)^{\mathrm{1}/\mathrm{3}} \left\{{a}^{\mathrm{1}/\mathrm{3}} +\left(−{a}\right)^{\mathrm{1}/\mathrm{3}} \right\} \\ $$$$\:\:=….. \\ $$$$ \\ $$
Commented by Rasheed.Sindhi last updated on 10/Dec/23
Is ((−a))^(1/6) =(−a)^(1/6) =a^(1/6) ? or (−a)^(1/3) =−a^(1/3) ?
$${Is}\:\sqrt[{\mathrm{6}}]{−{a}}\:=\left(−{a}\right)^{\mathrm{1}/\mathrm{6}} ={a}^{\mathrm{1}/\mathrm{6}} \:? \\ $$$${or}\:\left(−{a}\right)^{\mathrm{1}/\mathrm{3}} =−{a}^{\mathrm{1}/\mathrm{3}} ? \\ $$
Answered by mr W last updated on 10/Dec/23
let u=((1−(√(x^2 +1))))^(1/6) (((√(x^2 +1))−1))^(1/6) =(((−1)(1−(√(x^2 +1)))))^(1/6) =(i)^(1/3) u u+(i)^(1/3) u=1 u(1+(i)^(1/3) )=1 u=(1/(1+(i)^(1/3) ))=((1−(√(x^2 +1))))^(1/6) (√(x^2 +1))=1−(1/((1+(i)^(1/3) )^6 )) ⇒x=±(√([1−(1/((1+(i)^(1/3) )^6 ))]^2 −1))
$${let}\:{u}=\sqrt[{\mathrm{6}}]{\mathrm{1}−\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$\sqrt[{\mathrm{6}}]{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}−\mathrm{1}}=\sqrt[{\mathrm{6}}]{\left(−\mathrm{1}\right)\left(\mathrm{1}−\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\right)}=\sqrt[{\mathrm{3}}]{{i}}{u} \\ $$$${u}+\sqrt[{\mathrm{3}}]{{i}}{u}=\mathrm{1} \\ $$$${u}\left(\mathrm{1}+\sqrt[{\mathrm{3}}]{{i}}\right)=\mathrm{1} \\ $$$${u}=\frac{\mathrm{1}}{\mathrm{1}+\sqrt[{\mathrm{3}}]{{i}}}=\sqrt[{\mathrm{6}}]{\mathrm{1}−\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}=\mathrm{1}−\frac{\mathrm{1}}{\left(\mathrm{1}+\sqrt[{\mathrm{3}}]{{i}}\overset{\mathrm{6}} {\right)}} \\ $$$$\Rightarrow{x}=\pm\sqrt{\left[\mathrm{1}−\frac{\mathrm{1}}{\left(\mathrm{1}+\sqrt[{\mathrm{3}}]{{i}}\overset{\mathrm{6}} {\right)}}\right]^{\mathrm{2}} −\mathrm{1}} \\ $$
Answered by Frix last updated on 10/Dec/23
z^(1/6) +(−z)^(1/6) =1 z=re^(iθ) ∧−z=re^(i(θ−π)) z^(1/6) +(−z)^(1/6) ∈R ⇒ θ=−(θ−π) ⇒ θ=(π/2) ⇒ z=re^(i(π/2)) ∧−z=re^(−i(π/2)) (re^(i(π/2)) )^(1/6) +(re^(−i(π/2)) )^(1/6) =1 r^(1/6) (e^(i(π/(12))) +e^(−i(π/(12))) )=1 r^(1/6) (((√6)+(√2))/2)=1 r^(1/6) =(((√6)−(√2))/2) r=26−15(√3) z=(26−15(√3))i (26−15(√3))i=±(1−(√(x^2 +1))) This can be solved exactly...
$${z}^{\frac{\mathrm{1}}{\mathrm{6}}} +\left(−{z}\right)^{\frac{\mathrm{1}}{\mathrm{6}}} =\mathrm{1} \\ $$$${z}={r}\mathrm{e}^{\mathrm{i}\theta} \wedge−{z}={r}\mathrm{e}^{\mathrm{i}\left(\theta−\pi\right)} \\ $$$${z}^{\frac{\mathrm{1}}{\mathrm{6}}} +\left(−{z}\right)^{\frac{\mathrm{1}}{\mathrm{6}}} \in\mathbb{R}\:\Rightarrow\:\theta=−\left(\theta−\pi\right)\:\Rightarrow\:\theta=\frac{\pi}{\mathrm{2}} \\ $$$$\Rightarrow\:{z}={r}\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{2}}} \wedge−{z}={r}\mathrm{e}^{−\mathrm{i}\frac{\pi}{\mathrm{2}}} \\ $$$$\left({r}\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{2}}} \right)^{\frac{\mathrm{1}}{\mathrm{6}}} +\left({r}\mathrm{e}^{−\mathrm{i}\frac{\pi}{\mathrm{2}}} \right)^{\frac{\mathrm{1}}{\mathrm{6}}} =\mathrm{1} \\ $$$${r}^{\frac{\mathrm{1}}{\mathrm{6}}} \left(\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{12}}} +\mathrm{e}^{−\mathrm{i}\frac{\pi}{\mathrm{12}}} \right)=\mathrm{1} \\ $$$${r}^{\frac{\mathrm{1}}{\mathrm{6}}} \frac{\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}}{\mathrm{2}}=\mathrm{1} \\ $$$${r}^{\frac{\mathrm{1}}{\mathrm{6}}} =\frac{\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$${r}=\mathrm{26}−\mathrm{15}\sqrt{\mathrm{3}} \\ $$$${z}=\left(\mathrm{26}−\mathrm{15}\sqrt{\mathrm{3}}\right)\mathrm{i} \\ $$$$\left(\mathrm{26}−\mathrm{15}\sqrt{\mathrm{3}}\right)\mathrm{i}=\pm\left(\mathrm{1}−\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\right) \\ $$$$\mathrm{This}\:\mathrm{can}\:\mathrm{be}\:\mathrm{solved}\:\mathrm{exactly}… \\ $$
Answered by witcher3 last updated on 10/Dec/23
no real solution (1−(√(x^2 +1)))^(1/6) =e^((iπ)/6) ((√(x^2 +1))−1)^(1/6) +((√(x^2 +1))−1)^(1/6) =1 (√(x^2 +1))−1=((1/(e^((iπ)/6) +1)))^6 =(e^(−((iπ)/2)) /((2cos((π/(12))))^6 ))=−(i/(32cos^6 ((π/(12)))))=a x^2 +1=(a+1)^2 x^2 =a^2 +2a x=+_− (√(a^2 +2a)) a=−(i/(32cos^6 ((π/(12)))))
$$\mathrm{no}\:\mathrm{real}\:\mathrm{solution}\: \\ $$$$\left(\mathrm{1}−\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\right)^{\frac{\mathrm{1}}{\mathrm{6}}} =\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{6}}} \left(\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}−\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{6}}} +\left(\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}−\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{6}}} =\mathrm{1} \\ $$$$\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}−\mathrm{1}=\left(\frac{\mathrm{1}}{\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{6}}} +\mathrm{1}}\right)^{\mathrm{6}} =\frac{\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{2}}} }{\left(\mathrm{2cos}\left(\frac{\pi}{\mathrm{12}}\right)\right)^{\mathrm{6}} }=−\frac{\mathrm{i}}{\mathrm{32cos}^{\mathrm{6}} \left(\frac{\pi}{\mathrm{12}}\right)}=\mathrm{a} \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{1}=\left(\mathrm{a}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\mathrm{x}^{\mathrm{2}} =\mathrm{a}^{\mathrm{2}} +\mathrm{2a} \\ $$$$\mathrm{x}=\underset{−} {+}\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{2a}} \\ $$$$\mathrm{a}=−\frac{\mathrm{i}}{\mathrm{32cos}^{\mathrm{6}} \left(\frac{\pi}{\mathrm{12}}\right)} \\ $$

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