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Question Number 32371 by .none. last updated on 24/Mar/18

−1⟨x⟨0 (√x^2 )−(√((x+(1/x))^2 −4))=−2x+(1/x) Why?

$$−\mathrm{1}\langle{x}\langle\mathrm{0} \\ $$$$\sqrt{{x}^{\mathrm{2}} }−\sqrt{\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} −\mathrm{4}}=−\mathrm{2}{x}+\frac{\mathrm{1}}{{x}} \\ $$$${Why}? \\ $$

Commented by abdo imad last updated on 25/Mar/18

we have 0<−x<1 we use the ch.t=−x ⇒ 0<t<1 and prove that (√(t^2 )) −(√((t +(1/t))^2 −4)) =2t −(1/t) we have (√t^2 ) −(√((t+(1/t))^2 −4)) =t −(√(t^2 +2 +(1/t^2 )−4)) =t −(√((t−(1/t))^2 )) =t −∣t −(1/t)∣ but we have t −(1/t) =((t^2 −1)/t)<0 =t −((1/t) −t) =2t −(1/t) =−2x +(1/x) .

$${we}\:{have}\:\mathrm{0}<−{x}<\mathrm{1}\:\:{we}\:{use}\:{the}\:{ch}.{t}=−{x}\:\Rightarrow\:\mathrm{0}<{t}<\mathrm{1}\:{and} \\ $$$${prove}\:{that}\:\sqrt{{t}^{\mathrm{2}} \:}\:\:−\sqrt{\left({t}\:+\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} −\mathrm{4}}\:\:=\mathrm{2}{t}\:−\frac{\mathrm{1}}{{t}} \\ $$$${we}\:{have}\:\sqrt{{t}^{\mathrm{2}} }\:−\sqrt{\left({t}+\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} −\mathrm{4}}\: \\ $$$$={t}\:−\sqrt{{t}^{\mathrm{2}} \:+\mathrm{2}\:+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }−\mathrm{4}}\:={t}\:−\sqrt{\left({t}−\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} } \\ $$$$={t}\:−\mid{t}\:−\frac{\mathrm{1}}{{t}}\mid\:{but}\:{we}\:{have}\:{t}\:−\frac{\mathrm{1}}{{t}}\:=\frac{{t}^{\mathrm{2}} \:−\mathrm{1}}{{t}}<\mathrm{0} \\ $$$$={t}\:−\left(\frac{\mathrm{1}}{{t}}\:−{t}\right)\:=\mathrm{2}{t}\:−\frac{\mathrm{1}}{{t}}\:=−\mathrm{2}{x}\:+\frac{\mathrm{1}}{{x}}\:. \\ $$

Answered by MJS last updated on 24/Mar/18

I. (√x^2 )=−x with x<0 [i.e. (√((−2)^2 ))=(√4)=2] II. (√((x+(1/x))^2 −4))= [x≠0] =(√(x^2 +2+(1/x^2 )−4)) =(√(x^2 −2+(1/x^2 )))=(√((x−(1/x))^2 ))= =(x−(1/x)) with −1<x<0 [−1<x<0 ⇒ ∣x∣<1 ⇒ ∣(1/x)∣>1 ⇒ ⇒(x−(1/x))>0; i.e. −(1/2)−(1/(−(1/2)))= =−(1/2)+2=(3/2)] I.−II. −x−(x−(1/x))=−2x+(1/x)

$$\mathrm{I}. \\ $$$$\sqrt{{x}^{\mathrm{2}} }=−{x}\:\mathrm{with}\:{x}<\mathrm{0} \\ $$$$\:\:\:\:\:\left[\mathrm{i}.\mathrm{e}.\:\sqrt{\left(−\mathrm{2}\right)^{\mathrm{2}} }=\sqrt{\mathrm{4}}=\mathrm{2}\right] \\ $$$$ \\ $$$$\mathrm{II}. \\ $$$$\sqrt{\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} −\mathrm{4}}=\:\:\:\:\:\left[{x}\neq\mathrm{0}\right] \\ $$$$=\sqrt{{x}^{\mathrm{2}} +\mathrm{2}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\mathrm{4}} \\ $$$$=\sqrt{{x}^{\mathrm{2}} −\mathrm{2}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}=\sqrt{\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} }= \\ $$$$=\left({x}−\frac{\mathrm{1}}{{x}}\right)\:\mathrm{with}\:−\mathrm{1}<{x}<\mathrm{0} \\ $$$$\:\:\:\:\:\left[−\mathrm{1}<{x}<\mathrm{0}\:\Rightarrow\:\mid{x}\mid<\mathrm{1}\:\Rightarrow\:\mid\frac{\mathrm{1}}{{x}}\mid>\mathrm{1}\:\Rightarrow\right. \\ $$$$\:\:\:\:\:\Rightarrow\left({x}−\frac{\mathrm{1}}{{x}}\right)>\mathrm{0};\:\mathrm{i}.\mathrm{e}.\:−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{−\frac{\mathrm{1}}{\mathrm{2}}}= \\ $$$$\left.\:\:\:\:\:=−\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{2}=\frac{\mathrm{3}}{\mathrm{2}}\right] \\ $$$$ \\ $$$$\mathrm{I}.−\mathrm{II}. \\ $$$$−{x}−\left({x}−\frac{\mathrm{1}}{{x}}\right)=−\mathrm{2}{x}+\frac{\mathrm{1}}{{x}} \\ $$$$ \\ $$

Answered by $@ty@m last updated on 24/Mar/18

(√x^2 )=∣x∣=−x (∵−1<x<0) & (√((x+(1/x))^2 −4))=∣x−(1/x)∣=(1/x)−x

$$\sqrt{{x}^{\mathrm{2}} }=\mid{x}\mid=−{x}\:\left(\because−\mathrm{1}<{x}<\mathrm{0}\right) \\ $$$$\&\:\sqrt{\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} −\mathrm{4}}=\mid{x}−\frac{\mathrm{1}}{{x}}\mid=\frac{\mathrm{1}}{{x}}−{x} \\ $$

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