Question Number 225954 by fantastic2 last updated on 17/Nov/25
$$\sqrt{{x}−\frac{\mathrm{1}}{{x}}}+\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{x}}}={x} \\ $$$$ \\ $$
Answered by mr W last updated on 17/Nov/25
Commented by mr W last updated on 17/Nov/25
$${we}\:{can}\:{construct}\:{a}\:{triangle}\:{as} \\ $$$${shown}. \\ $$$${AD}=\sqrt{\left(\sqrt{{x}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{1}}{\:\sqrt{{x}}}\right)^{\mathrm{2}} }=\sqrt{{x}−\frac{\mathrm{1}}{{x}}} \\ $$$${DC}=\sqrt{\mathrm{1}^{\mathrm{2}} −\left(\frac{\mathrm{1}}{\:\sqrt{{x}}}\right)^{\mathrm{2}} }=\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{x}}} \\ $$$${AC}=\sqrt{{x}−\frac{\mathrm{1}}{{x}}}+\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{x}}}={x} \\ $$$${now}\:{let}'{s}\:{look}\:{at}\:{triangle}'{s}\:{area}\:\:. \\ $$$$\:=\frac{{AC}×{BD}}{\mathrm{2}}=\frac{{x}×\frac{\mathrm{1}}{\:\sqrt{{x}}}}{\mathrm{2}}=\frac{\sqrt{{x}}}{\mathrm{2}} \\ $$$${on}\:{the}\:{other}\:{side}, \\ $$$$\:=\frac{\mathrm{1}×\sqrt{{x}}×\mathrm{sin}\:\angle{B}}{\mathrm{2}}=\frac{\sqrt{{x}}\:\mathrm{sin}\:\angle{B}}{\mathrm{2}}=\frac{\sqrt{{x}}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{sin}\:\angle{B}=\mathrm{1}\:\Rightarrow\angle{B}=\mathrm{90}°\: \\ $$$$\Rightarrow{ABC}\:{is}\:{right}−{angled}\:{triangle} \\ $$$${x}^{\mathrm{2}} =\left(\sqrt{{x}}\right)^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} −{x}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{x}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:\:\:\:\:\:\:\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}<\mathrm{0}\:{rejected}\right) \\ $$
Commented by fantastic2 last updated on 17/Nov/25
$${you}\:{did}\:{this}\:{in}\:{the}\:{exact}\:{same} \\ $$$${method}\:{i}\:{expected}. \\ $$$${great}\:{sir}! \\ $$
Answered by Frix last updated on 17/Nov/25
$${x}=\varphi=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\mathrm{We}\:\mathrm{only}\:\mathrm{need}\:\mathrm{this}\:\mathrm{identity}:\:\frac{\mathrm{1}}{\varphi}=\varphi−\mathrm{1} \\ $$$$\Rightarrow \\ $$$$\sqrt{\varphi−\frac{\mathrm{1}}{\varphi}}=\mathrm{1} \\ $$$$\sqrt{\mathrm{1}−\frac{\mathrm{1}}{\varphi}}=\sqrt{\frac{\varphi−\mathrm{1}}{\varphi}}=\frac{\mathrm{1}}{\varphi} \\ $$$$\mathrm{1}+\frac{\mathrm{1}}{\varphi}=\varphi \\ $$
Commented by fantastic2 last updated on 17/Nov/25
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Answered by Kademi last updated on 18/Nov/25
$$\:\sqrt{{x}−\frac{\mathrm{1}}{{x}}}\:+\:\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{x}}}\:=\:{x} \\ $$$$\:\sqrt{{x}^{\mathrm{3}} −{x}}\:+\:\sqrt{{x}^{\mathrm{2}} −{x}}\:=\:{x}^{\mathrm{2}} \\ $$$$\:\left(\sqrt{{x}^{\mathrm{3}} −{x}}\right)^{\mathrm{2}} \:=\:\left({x}^{\mathrm{2}} −\sqrt{{x}^{\mathrm{2}} −{x}}\right)^{\mathrm{2}} \\ $$$$\:{x}^{\mathrm{2}} −{x}+\mathrm{1}\:=\:\mathrm{2}\sqrt{{x}^{\mathrm{2}} −{x}} \\ $$$$\:\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{\mathrm{2}} \:=\:\left(\mathrm{2}\sqrt{{x}^{\mathrm{2}} −{x}}\right)^{\mathrm{2}} \\ $$$$\:{x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{3}} −{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}\:=\:\mathrm{0} \\ $$$$\:\left({x}^{\mathrm{2}} −{x}+\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\sqrt{\mathrm{5}}\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)\right)^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\:\begin{cases}{{x}^{\mathrm{2}} +\left(−\mathrm{1}−\sqrt{\mathrm{5}}\right){x}+\left(\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\:=\:\mathrm{0}}\\{{x}^{\mathrm{2}} +\left(\mathrm{1}+\sqrt{\mathrm{5}}\right){x}+\left(\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\:=\:\mathrm{0}}\end{cases} \\ $$$$\:{x}_{\mathrm{1},\mathrm{2}} \:=\:\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:=\:\varphi \\ $$$$\:{x}_{\mathrm{3},\mathrm{4}} \:=\:\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\:=\:\mathrm{1}−\varphi \\ $$$$\sqrt{{x}−\frac{\mathrm{1}}{{x}}}\:+\:\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{x}}}\:=\:{x}\:\:\rightarrow\:\:\therefore\:\varphi+\mathrm{1}\:\neq\:\varphi \\ $$$$\:\because\:{x}\:=\:\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:=\:\varphi \\ $$