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Question Number 147893 by mathdanisur last updated on 24/Jul/21

x + 1 + (1/x) = 4  ⇒  x + 1 - (1/x) = ?

$${x}\:+\:\mathrm{1}\:+\:\frac{\mathrm{1}}{{x}}\:=\:\mathrm{4}\:\:\Rightarrow\:\:{x}\:+\:\mathrm{1}\:-\:\frac{\mathrm{1}}{{x}}\:=\:? \\ $$

Commented by aliibrahim1 last updated on 24/Jul/21

x+(1/x)=3⇛     (x+(1/x))^2 =3^2   x^2 +(1/x^2 )+2=9  x^2 +(1/x^2 )−2=5  (x−(1/x))^2 =5 ⇛     x−(1/x)=(√5)  x−(1/x)+1=(√5)+1

$${x}+\frac{\mathrm{1}}{{x}}=\mathrm{3}\Rrightarrow\:\:\:\:\:\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} =\mathrm{3}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\mathrm{2}=\mathrm{9} \\ $$$${x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\mathrm{2}=\mathrm{5} \\ $$$$\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} =\mathrm{5}\:\Rrightarrow\:\:\:\:\:{x}−\frac{\mathrm{1}}{{x}}=\sqrt{\mathrm{5}} \\ $$$${x}−\frac{\mathrm{1}}{{x}}+\mathrm{1}=\sqrt{\mathrm{5}}+\mathrm{1} \\ $$

Commented by mathdanisur last updated on 24/Jul/21

Thankyou Sir

$${Thankyou}\:{Sir} \\ $$

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