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Question Number 200498 by mr W last updated on 19/Nov/23
solve for x∈R (√(x−(1/x)))+(√(1−(1/x)))=x
$${solve}\:{for}\:{x}\in{R} \\ $$$$\sqrt{{x}−\frac{\mathrm{1}}{{x}}}+\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{x}}}={x} \\ $$
Commented by BaliramKumar last updated on 19/Nov/23
Q. 192958
$$\mathrm{Q}.\:\mathrm{192958} \\ $$
Commented by mr W last updated on 19/Nov/23
reposted for alternative solutions.
$${reposted}\:{for}\:{alternative}\:{solutions}. \\ $$
Commented by Harnada last updated on 20/Nov/23
Q183839
$${Q}\mathrm{183839} \\ $$
Answered by cortano12 last updated on 19/Nov/23
$$\:\:\cancel{\underline{\underbrace{\:}}} \\ $$
Answered by Rasheed.Sindhi last updated on 19/Nov/23
(√(x−(1/x)))+(√(1−(1/x))) =x.........(i) (x−(1/x))−(1−(1/x))=x((√(x−(1/x))) −(√(1−(1/x))) ) (√(x−(1/x))) −(√(1−(1/x))) =((x−1)/x)...(ii) (i)+(ii): 2(√(x−(1/x))) =x−(1/x)+1 (√(x−(1/x))) =y (say) 2y=y^2 +1 y^2 −2y+1=0 (y−1)^2 =0 y=1 x−(1/x)=1 x^2 −x−1=0 x_(≥0) =((1+(√(1+4)))/2)=((1+(√5))/2)
$$\sqrt{{x}−\frac{\mathrm{1}}{{x}}}+\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{x}}}\:={x}………\left({i}\right) \\ $$$$\left({x}−\frac{\mathrm{1}}{{x}}\right)−\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)={x}\left(\sqrt{{x}−\frac{\mathrm{1}}{{x}}}\:−\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{x}}}\:\right) \\ $$$$\sqrt{{x}−\frac{\mathrm{1}}{{x}}}\:−\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{x}}}\:=\frac{{x}−\mathrm{1}}{{x}}…\left({ii}\right) \\ $$$$\left({i}\right)+\left({ii}\right): \\ $$$$\mathrm{2}\sqrt{{x}−\frac{\mathrm{1}}{{x}}}\:={x}−\frac{\mathrm{1}}{{x}}+\mathrm{1} \\ $$$$\sqrt{{x}−\frac{\mathrm{1}}{{x}}}\:={y}\:\left({say}\right) \\ $$$$\mathrm{2}{y}={y}^{\mathrm{2}} +\mathrm{1} \\ $$$${y}^{\mathrm{2}} −\mathrm{2}{y}+\mathrm{1}=\mathrm{0} \\ $$$$\left({y}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${y}=\mathrm{1} \\ $$$${x}−\frac{\mathrm{1}}{{x}}=\mathrm{1} \\ $$$${x}^{\mathrm{2}} −{x}−\mathrm{1}=\mathrm{0} \\ $$$${x}_{\geqslant\mathrm{0}} =\frac{\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{4}}}{\mathrm{2}}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$
Answered by Rasheed.Sindhi last updated on 19/Nov/23
a+b=x.........(i) a^2 −b^2 =(x−(1/x))−(1−(1/x))=x−1 a−b=((a^2 −b^2 )/(a+b))=((x−1)/x)......(ii) (i)+(ii): 2a=x+((x−1)/x)=x−(1/x)+1 2((√(x−(1/x)))_(y) )=x−(1/x)+1 2y=y^2 +1 (y−1)^2 =0 y=1 (√(x−(1/x))) =1 x−(1/x)=1 x^2 −x−1=0 x_(≥0) =((1+(√5) )/2)
$$\:\:\:\:{a}+{b}={x}………\left({i}\right) \\ $$$${a}^{\mathrm{2}} −{b}^{\mathrm{2}} =\left({x}−\frac{\mathrm{1}}{{x}}\right)−\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)={x}−\mathrm{1} \\ $$$${a}−{b}=\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{a}+{b}}=\frac{{x}−\mathrm{1}}{{x}}……\left({ii}\right) \\ $$$$\left({i}\right)+\left({ii}\right): \\ $$$$\mathrm{2}{a}={x}+\frac{{x}−\mathrm{1}}{{x}}={x}−\frac{\mathrm{1}}{{x}}+\mathrm{1} \\ $$$$\mathrm{2}\left(\underset{{y}} {\underbrace{\sqrt{{x}−\frac{\mathrm{1}}{{x}}}}}\right)={x}−\frac{\mathrm{1}}{{x}}+\mathrm{1} \\ $$$$\mathrm{2}{y}={y}^{\mathrm{2}} +\mathrm{1} \\ $$$$\left({y}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${y}=\mathrm{1} \\ $$$$\sqrt{{x}−\frac{\mathrm{1}}{{x}}}\:=\mathrm{1} \\ $$$${x}−\frac{\mathrm{1}}{{x}}=\mathrm{1} \\ $$$${x}^{\mathrm{2}} −{x}−\mathrm{1}=\mathrm{0} \\ $$$${x}_{\geqslant\mathrm{0}} =\frac{\mathrm{1}+\sqrt{\mathrm{5}}\:}{\mathrm{2}} \\ $$
Answered by Frix last updated on 19/Nov/23
(√(x−(1/x)))=u∧(√(1−(1/x)))=v ⇒ x=u+v; u, v ≥0 (1) u+v−(1/(u+v))=u^2 ⇒ u^3 +u^2 (v−1)−2uv−v^2 +1=0 (2) 1−(1/(u+v))=v^2 ⇒ u=((v^3 −v+1)/(1−v^2 )) (1) (((v^2 +v−1)^2 )/((1−v^2 )^3 ))=0 ⇒ v=−(1/2)+((√5)/2) ⇒ u=1 x=u+v=(1/2)+((√5)/2)
$$\sqrt{{x}−\frac{\mathrm{1}}{{x}}}={u}\wedge\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{x}}}={v}\:\Rightarrow\:{x}={u}+{v};\:{u},\:{v}\:\geqslant\mathrm{0} \\ $$$$\left(\mathrm{1}\right)\:{u}+{v}−\frac{\mathrm{1}}{{u}+{v}}={u}^{\mathrm{2}} \:\Rightarrow\:{u}^{\mathrm{3}} +{u}^{\mathrm{2}} \left({v}−\mathrm{1}\right)−\mathrm{2}{uv}−{v}^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{1}−\frac{\mathrm{1}}{{u}+{v}}={v}^{\mathrm{2}} \:\Rightarrow\:{u}=\frac{{v}^{\mathrm{3}} −{v}+\mathrm{1}}{\mathrm{1}−{v}^{\mathrm{2}} } \\ $$$$\left(\mathrm{1}\right)\:\frac{\left({v}^{\mathrm{2}} +{v}−\mathrm{1}\right)^{\mathrm{2}} }{\left(\mathrm{1}−{v}^{\mathrm{2}} \right)^{\mathrm{3}} }=\mathrm{0}\:\Rightarrow\:{v}=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\:\Rightarrow\:{u}=\mathrm{1} \\ $$$${x}={u}+{v}=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$
Answered by mr W last updated on 19/Nov/23
Commented by Rasheed.Sindhi last updated on 20/Nov/23
Great! Thanks sir!
$$\mathbb{G}\boldsymbol{\mathrm{reat}}! \\ $$$$\:\boldsymbol{\mathcal{T}{hanks}}\:\boldsymbol{{sir}}! \\ $$
Commented by mr W last updated on 19/Nov/23
x>0 ((√x))^2 +1^2 =x^2 x^2 −x−1=0 ⇒x=((1+(√5))/2)
$${x}>\mathrm{0} \\ $$$$\left(\sqrt{{x}}\right)^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} ={x}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} −{x}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{x}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$
Commented by Rasheed.Sindhi last updated on 20/Nov/23
You see geometry in algebra where others do not see!
$${You}\:{see}\:{geometry}\:{in}\:{algebra} \\ $$$$\boldsymbol{{where}}\:{others}\:{do}\:{not}\:{see}! \\ $$
Commented by Rasheed.Sindhi last updated on 20/Nov/23
But sir I can′t understand how is the answer related to the question?
$$\mathcal{B}{ut}\:{sir}\:{I}\:{can}'{t}\:{understand}\:{how}\:{is} \\ $$$${the}\:{answer}\:{related}\:{to}\:{the}\:{question}? \\ $$
Commented by manolex last updated on 20/Nov/23
ingredients: (√(x−(1/x) )),(√(1−(1/x))) ,x product triangle 1,x,(√x),h=(1/( (√x))) way sustentation .saludos mr W
$$ \\ $$$${ingredients}:\:\:\:\:\sqrt{{x}−\frac{\mathrm{1}}{{x}}\:},\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{x}}}\:,{x}\:\:\:\:\:{product}\:\:\:\:{triangle}\:\:\:\:\mathrm{1},{x},\sqrt{{x}},{h}=\frac{\mathrm{1}}{\:\sqrt{{x}}}\:\:\:\:\:{way}\:{sustentation}\:.{saludos}\:{mr}\:\:{W} \\ $$
Commented by mr W last updated on 20/Nov/23
thanks you sir! to be honest, i have a fondness for geometry. at first i always try to understand if there is a geometrical meaning or interpretation for the question. in this way i have solved some questions in this forum which at the first glance have nothing to do with geometry.
$${thanks}\:{you}\:{sir}! \\ $$$${to}\:{be}\:{honest},\:{i}\:{have}\:{a}\:{fondness}\:{for} \\ $$$${geometry}.\:{at}\:{first}\:{i}\:{always}\:{try}\:{to}\: \\ $$$${understand}\:{if}\:{there}\:{is}\:{a}\:{geometrical} \\ $$$${meaning}\:{or}\:{interpretation}\:{for}\:{the} \\ $$$${question}.\:{in}\:{this}\:{way}\:{i}\:{have}\:{solved}\: \\ $$$${some}\:{questions}\:{in}\:{this}\:{forum}\:{which} \\ $$$${at}\:{the}\:{first}\:{glance}\:{have}\:{nothing}\:{to}\:{do} \\ $$$${with}\:{geometry}. \\ $$
Commented by mr W last updated on 20/Nov/23
Commented by mr W last updated on 20/Nov/23
BD=(√(((√x))^2 −((1/( (√x))))^2 ))=(√(x−(1/x))) DC=(√(1^2 −((1/( (√x))))^2 ))=(√(1−(1/x))) BC=(√(x−(1/x)))+(√(1−(1/x)))=x (as given) ((AB)/(AC))=((√x)/1)=(√x)=((BC)/(AB))=(x/( (√x)))=(√x)=((AC)/(AD))=(1/(1/( (√x))))=(√x) ⇒ΔABC is right−angled triangle ⇒((√x))^2 +1^2 =x^2 ⇒x^2 −x−1=0
$${BD}=\sqrt{\left(\sqrt{{x}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{1}}{\:\sqrt{{x}}}\right)^{\mathrm{2}} }=\sqrt{{x}−\frac{\mathrm{1}}{{x}}} \\ $$$${DC}=\sqrt{\mathrm{1}^{\mathrm{2}} −\left(\frac{\mathrm{1}}{\:\sqrt{{x}}}\right)^{\mathrm{2}} }=\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{x}}} \\ $$$${BC}=\sqrt{{x}−\frac{\mathrm{1}}{{x}}}+\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{x}}}={x}\:\:\:\left({as}\:{given}\right) \\ $$$$\frac{{AB}}{{AC}}=\frac{\sqrt{{x}}}{\mathrm{1}}=\sqrt{{x}}=\frac{{BC}}{{AB}}=\frac{{x}}{\:\sqrt{{x}}}=\sqrt{{x}}=\frac{{AC}}{{AD}}=\frac{\mathrm{1}}{\frac{\mathrm{1}}{\:\sqrt{{x}}}}=\sqrt{{x}} \\ $$$$\Rightarrow\Delta{ABC}\:{is}\:{right}−{angled}\:{triangle} \\ $$$$\Rightarrow\left(\sqrt{{x}}\right)^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} ={x}^{\mathrm{2}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} −{x}−\mathrm{1}=\mathrm{0} \\ $$
Commented by necx122 last updated on 20/Nov/23
I've always suspected the same that you and Ajfour had a special likeness for geometry because your manner of approach tackling geometry based questions is most often overwhelming.

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