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Question-63095

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Question Number 63095 by aliesam last updated on 28/Jun/19
Answered by MJS last updated on 29/Jun/19
((4x^4 −40x^2 +100))^(1/3) −((2x^2 −10))^(1/3) =20−2x^2 x^2 =s ((4s^2 −40s+100))^(1/3) −((2s−10))^(1/3) =20−2s ((4(s−5)^2 ))^(1/3) −((2(s−5)))^(1/3) =10−2(s−5) t=2(s−5) ⇔ s=(t/2)+5 (t^2 )^(1/3) −(t)^(1/3) =10−t u=(t)^(1/3) ⇔ t=u^3 u^2 −u=10−u^3 (u−2)(u^2 +3u+5)=0 u_1 =2 ⇒ t_1 =8 ⇒ s_1 =9 ⇒ x=±3 u_2 =−(3/2)−((√(11))/2)i ⇒ t_2 =9−2(√(11))i not valid u_3 =−(3/2)+((√(11))/2)i ⇒ t_3 =9+2(√(11))i not valid t_2 and t_3 do not solve (t^2 )^(1/3) −(t)^(1/3) =10−t
$$\sqrt[{\mathrm{3}}]{\mathrm{4}{x}^{\mathrm{4}} −\mathrm{40}{x}^{\mathrm{2}} +\mathrm{100}}−\sqrt[{\mathrm{3}}]{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{10}}=\mathrm{20}−\mathrm{2}{x}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} ={s} \\ $$$$\sqrt[{\mathrm{3}}]{\mathrm{4}{s}^{\mathrm{2}} −\mathrm{40}{s}+\mathrm{100}}−\sqrt[{\mathrm{3}}]{\mathrm{2}{s}−\mathrm{10}}=\mathrm{20}−\mathrm{2}{s} \\ $$$$\sqrt[{\mathrm{3}}]{\mathrm{4}\left({s}−\mathrm{5}\right)^{\mathrm{2}} }−\sqrt[{\mathrm{3}}]{\mathrm{2}\left({s}−\mathrm{5}\right)}=\mathrm{10}−\mathrm{2}\left({s}−\mathrm{5}\right) \\ $$$${t}=\mathrm{2}\left({s}−\mathrm{5}\right)\:\Leftrightarrow\:{s}=\frac{{t}}{\mathrm{2}}+\mathrm{5} \\ $$$$\sqrt[{\mathrm{3}}]{{t}^{\mathrm{2}} }−\sqrt[{\mathrm{3}}]{{t}}=\mathrm{10}−{t} \\ $$$${u}=\sqrt[{\mathrm{3}}]{{t}}\:\Leftrightarrow\:{t}={u}^{\mathrm{3}} \\ $$$${u}^{\mathrm{2}} −{u}=\mathrm{10}−{u}^{\mathrm{3}} \\ $$$$\left({u}−\mathrm{2}\right)\left({u}^{\mathrm{2}} +\mathrm{3}{u}+\mathrm{5}\right)=\mathrm{0} \\ $$$${u}_{\mathrm{1}} =\mathrm{2}\:\Rightarrow\:{t}_{\mathrm{1}} =\mathrm{8}\:\Rightarrow\:{s}_{\mathrm{1}} =\mathrm{9}\:\Rightarrow\:{x}=\pm\mathrm{3} \\ $$$${u}_{\mathrm{2}} =−\frac{\mathrm{3}}{\mathrm{2}}−\frac{\sqrt{\mathrm{11}}}{\mathrm{2}}\mathrm{i}\:\Rightarrow\:{t}_{\mathrm{2}} =\mathrm{9}−\mathrm{2}\sqrt{\mathrm{11}}\mathrm{i}\:\mathrm{not}\:\mathrm{valid} \\ $$$${u}_{\mathrm{3}} =−\frac{\mathrm{3}}{\mathrm{2}}+\frac{\sqrt{\mathrm{11}}}{\mathrm{2}}\mathrm{i}\:\Rightarrow\:{t}_{\mathrm{3}} =\mathrm{9}+\mathrm{2}\sqrt{\mathrm{11}}\mathrm{i}\:\mathrm{not}\:\mathrm{valid} \\ $$$${t}_{\mathrm{2}} \:\mathrm{and}\:{t}_{\mathrm{3}} \:\mathrm{do}\:\mathrm{not}\:\mathrm{solve}\:\sqrt[{\mathrm{3}}]{{t}^{\mathrm{2}} }−\sqrt[{\mathrm{3}}]{{t}}=\mathrm{10}−{t} \\ $$
Commented by MJS last updated on 29/Jun/19
if we write t_2 and t_3 as re^(±iθ) and add ±2π then re^(±i(θ+2π)) solve the equation but I doubt if these solution are legit...
$$\mathrm{if}\:\mathrm{we}\:\mathrm{write}\:{t}_{\mathrm{2}} \:\mathrm{and}\:{t}_{\mathrm{3}} \:\mathrm{as}\:{r}\mathrm{e}^{\pm\mathrm{i}\theta} \:\mathrm{and}\:\mathrm{add}\:\pm\mathrm{2}\pi \\ $$$$\mathrm{then}\:{r}\mathrm{e}^{\pm\mathrm{i}\left(\theta+\mathrm{2}\pi\right)} \:\mathrm{solve}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{but}\:\mathrm{I}\:\mathrm{doubt} \\ $$$$\mathrm{if}\:\mathrm{these}\:\mathrm{solution}\:\mathrm{are}\:\mathrm{legit}… \\ $$

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