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Question-164549

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Question Number 164549 by wwww last updated on 18/Jan/22
Answered by alephzero last updated on 18/Jan/22
(a) a−(1/a)=4 ((a^2 −1)/a) = 4 ((a^2 −1−4a)/a) = 0 ⇒ a^2 −4a−1 = 0 a = ((4 ± (√(16+4)))/2) = ((4 ± (√(20)))/2) = ((4 ± 2(√5))/2) = = 2 ± (√5) Let a+(1/a) =A ⇒ A_1 = 2+(√5)+(1/(2+(√5)))=2+(√5)−2+(√5) = = 2(√5) A_2 = 2−(√5)+(1/(2−(√5))) =2−(√5)−2−(√5) = = −2(√5) ⇒ a+(1/a) = ±2(√5) (b) a^3 +(1/a^3 ) = B a_1 ^3 = 38+17(√5) ∧ a_2 ^3 = 38−17(√5) ⇒ B_1 = 38+17(√5)+(1/(38+17(√5))) = 34(√5) B_2 = 38−17(√5)+(1/(38−17(√5))) = −34(√5)
$$\left({a}\right)\:{a}−\frac{\mathrm{1}}{{a}}=\mathrm{4} \\ $$$$\frac{{a}^{\mathrm{2}} −\mathrm{1}}{{a}}\:=\:\mathrm{4} \\ $$$$\frac{{a}^{\mathrm{2}} −\mathrm{1}−\mathrm{4}{a}}{{a}}\:=\:\mathrm{0} \\ $$$$\Rightarrow\:{a}^{\mathrm{2}} −\mathrm{4}{a}−\mathrm{1}\:=\:\mathrm{0} \\ $$$${a}\:=\:\frac{\mathrm{4}\:\pm\:\sqrt{\mathrm{16}+\mathrm{4}}}{\mathrm{2}}\:=\:\frac{\mathrm{4}\:\pm\:\sqrt{\mathrm{20}}}{\mathrm{2}}\:=\:\frac{\mathrm{4}\:\pm\:\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{2}}\:= \\ $$$$=\:\mathrm{2}\:\pm\:\sqrt{\mathrm{5}} \\ $$$$\mathrm{Let}\:{a}+\frac{\mathrm{1}}{{a}}\:=\mathcal{A} \\ $$$$\Rightarrow\:\mathcal{A}_{\mathrm{1}} \:=\:\mathrm{2}+\sqrt{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{2}+\sqrt{\mathrm{5}}}=\mathrm{2}+\sqrt{\mathrm{5}}−\mathrm{2}+\sqrt{\mathrm{5}}\:= \\ $$$$=\:\mathrm{2}\sqrt{\mathrm{5}} \\ $$$$\mathcal{A}_{\mathrm{2}} \:=\:\mathrm{2}−\sqrt{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{2}−\sqrt{\mathrm{5}}}\:=\mathrm{2}−\sqrt{\mathrm{5}}−\mathrm{2}−\sqrt{\mathrm{5}}\:= \\ $$$$=\:−\mathrm{2}\sqrt{\mathrm{5}} \\ $$$$\Rightarrow\:{a}+\frac{\mathrm{1}}{{a}}\:=\:\pm\mathrm{2}\sqrt{\mathrm{5}} \\ $$$$\left({b}\right)\:{a}^{\mathrm{3}} +\frac{\mathrm{1}}{{a}^{\mathrm{3}} }\:=\:\mathcal{B} \\ $$$${a}_{\mathrm{1}} ^{\mathrm{3}} \:=\:\mathrm{38}+\mathrm{17}\sqrt{\mathrm{5}}\:\wedge\:{a}_{\mathrm{2}} ^{\mathrm{3}} \:=\:\mathrm{38}−\mathrm{17}\sqrt{\mathrm{5}} \\ $$$$\Rightarrow\:\mathcal{B}_{\mathrm{1}} \:=\:\mathrm{38}+\mathrm{17}\sqrt{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{38}+\mathrm{17}\sqrt{\mathrm{5}}}\:=\:\mathrm{34}\sqrt{\mathrm{5}} \\ $$$$\mathcal{B}_{\mathrm{2}} \:=\:\mathrm{38}−\mathrm{17}\sqrt{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{38}−\mathrm{17}\sqrt{\mathrm{5}}}\:=\:−\mathrm{34}\sqrt{\mathrm{5}} \\ $$
Answered by Rasheed.Sindhi last updated on 18/Jan/22
a−(1/a)=4 (a): (a−(1/a)=4)^2 a^2 +(1/a^2 )−2=16 (a+(1/a))^2 =20 a+(1/a)=±2(√5) (b):(a+(1/a))^3 =(±2(√5))^3 a^3 +(1/a^3 )+3(a+(1/a))=±40(√5) a^3 +(1/a^3 )+3(±2(√5))=±40(√5) a^3 +(1/a^3 )=±40(√5) ∓6(√5) =±34(√5) (c):( a−(1/a)=4)^3 a^3 −(1/a^3 )−3(a−(1/a))=64 a^3 −(1/a^3 )−3(4)=64 a^3 −(1/a^3 )=76 (a^3 −(1/a^3 ))(a^3 +(1/a^3 ))=(76)(±34(√5)) a^6 −(1/a^6 )=±2584(√5)
$${a}−\frac{\mathrm{1}}{{a}}=\mathrm{4} \\ $$$$\left({a}\right):\:\:\left({a}−\frac{\mathrm{1}}{{a}}=\mathrm{4}\right)^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} +\frac{\mathrm{1}}{{a}^{\mathrm{2}} }−\mathrm{2}=\mathrm{16} \\ $$$$\left({a}+\frac{\mathrm{1}}{{a}}\right)^{\mathrm{2}} =\mathrm{20} \\ $$$${a}+\frac{\mathrm{1}}{{a}}=\pm\mathrm{2}\sqrt{\mathrm{5}} \\ $$$$\left({b}\right):\left({a}+\frac{\mathrm{1}}{{a}}\right)^{\mathrm{3}} =\left(\pm\mathrm{2}\sqrt{\mathrm{5}}\right)^{\mathrm{3}} \\ $$$${a}^{\mathrm{3}} +\frac{\mathrm{1}}{{a}^{\mathrm{3}} }+\mathrm{3}\left({a}+\frac{\mathrm{1}}{{a}}\right)=\pm\mathrm{40}\sqrt{\mathrm{5}} \\ $$$${a}^{\mathrm{3}} +\frac{\mathrm{1}}{{a}^{\mathrm{3}} }+\mathrm{3}\left(\pm\mathrm{2}\sqrt{\mathrm{5}}\right)=\pm\mathrm{40}\sqrt{\mathrm{5}} \\ $$$${a}^{\mathrm{3}} +\frac{\mathrm{1}}{{a}^{\mathrm{3}} }=\pm\mathrm{40}\sqrt{\mathrm{5}}\:\mp\mathrm{6}\sqrt{\mathrm{5}}\:=\pm\mathrm{34}\sqrt{\mathrm{5}} \\ $$$$\left({c}\right):\left(\:{a}−\frac{\mathrm{1}}{{a}}=\mathrm{4}\right)^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:{a}^{\mathrm{3}} −\frac{\mathrm{1}}{{a}^{\mathrm{3}} }−\mathrm{3}\left({a}−\frac{\mathrm{1}}{{a}}\right)=\mathrm{64} \\ $$$$\:\:\:\:\:\:\:\:{a}^{\mathrm{3}} −\frac{\mathrm{1}}{{a}^{\mathrm{3}} }−\mathrm{3}\left(\mathrm{4}\right)=\mathrm{64} \\ $$$$\:\:\:\:\:\:{a}^{\mathrm{3}} −\frac{\mathrm{1}}{{a}^{\mathrm{3}} }=\mathrm{76} \\ $$$$\:\:\:\:\left({a}^{\mathrm{3}} −\frac{\mathrm{1}}{{a}^{\mathrm{3}} }\right)\left({a}^{\mathrm{3}} +\frac{\mathrm{1}}{{a}^{\mathrm{3}} }\right)=\left(\mathrm{76}\right)\left(\pm\mathrm{34}\sqrt{\mathrm{5}}\right) \\ $$$$\:\:\:\:{a}^{\mathrm{6}} −\frac{\mathrm{1}}{{a}^{\mathrm{6}} }=\pm\mathrm{2584}\sqrt{\mathrm{5}} \\ $$$$ \\ $$

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