Question-82593

Question Number 82593 by Power last updated on 22/Feb/20

Commented by mr W last updated on 23/Feb/20

t=(√x)>0 t^4 −16t−12=0 (t^2 −2t−2)(t^2 +2t+6)=0 t^2 +2t+6=(t+1)^2 +5>0 t^2 −2t−2=0 ⇒t^2 −2t=2=x−2(√x) if x∈C: t=−1±(√5)i x−2(√x)=t^2 −2t=t^2 +2t−4t =−6−4t =−6−4(−1±(√5)i) =−2±4(√5)i

$${t}=\sqrt{{x}}>\mathrm{0} \\ $$$${t}^{\mathrm{4}} −\mathrm{16}{t}−\mathrm{12}=\mathrm{0} \\ $$$$\left({t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{2}\right)\left({t}^{\mathrm{2}} +\mathrm{2}{t}+\mathrm{6}\right)=\mathrm{0} \\ $$$${t}^{\mathrm{2}} +\mathrm{2}{t}+\mathrm{6}=\left({t}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{5}>\mathrm{0} \\ $$$${t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{2}=\mathrm{0}\: \\ $$$$\Rightarrow{t}^{\mathrm{2}} −\mathrm{2}{t}=\mathrm{2}={x}−\mathrm{2}\sqrt{{x}} \\ $$$$ \\ $$$${if}\:{x}\in\mathbb{C}: \\ $$$${t}=−\mathrm{1}\pm\sqrt{\mathrm{5}}{i}\: \\ $$$${x}−\mathrm{2}\sqrt{{x}}={t}^{\mathrm{2}} −\mathrm{2}{t}={t}^{\mathrm{2}} +\mathrm{2}{t}−\mathrm{4}{t} \\ $$$$=−\mathrm{6}−\mathrm{4}{t} \\ $$$$=−\mathrm{6}−\mathrm{4}\left(−\mathrm{1}\pm\sqrt{\mathrm{5}}{i}\right) \\ $$$$=−\mathrm{2}\pm\mathrm{4}\sqrt{\mathrm{5}}{i} \\ $$

Commented by peter frank last updated on 23/Feb/20

$$\:{mr}\:\:{w}\:{please}\:{help}\:\mathrm{82596} \\ $$

Commented by Power last updated on 23/Feb/20

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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