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Question Number 212647 by golsendro last updated on 20/Oct/24

$$\sqrt{\mathrm{x}-\frac{1}{\mathrm{x}}}+\sqrt{1-\frac{1}{\mathrm{x}}}=\mathrm{x}$$

Commented by Rasheed.Sindhi last updated on 21/Oct/24

$$\text{You can't use 'macro parameter character \#' in math mode}$$

Answered by Frix last updated on 20/Oct/24

$$x=\phi =\frac{1+\sqrt{5}}{2}$$$$\frac{1}{\phi }=\phi -1\iff \phi =\frac{1}{\phi }+1\iff \phi -\frac{1}{\phi }=1$$$$1-\frac{1}{\phi }=\frac{\phi -1}{\phi }=\frac{1}{\phi }(\phi -1)=\frac{1}{{\phi }^2}$$$$\sqrt{\phi -\frac{1}{\phi }}+\sqrt{1-\frac{1}{\phi }}=1+\frac{1}{\phi }=\phi$$

Answered by MATHEMATICSAM last updated on 22/Oct/24

$$\sqrt{\mathrm{x}-\frac{1}{\mathrm{x}}}+\sqrt{1-\frac{1}{\mathrm{x}}}=\mathrm{x}....\left(\mathrm{i}\right)$$$$\Rightarrow (\sqrt{\mathrm{x}-\frac{1}{\mathrm{x}}}+\sqrt{1-\frac{1}{\mathrm{x}}})\times$$$$(\sqrt{\mathrm{x}-\frac{1}{\mathrm{x}}}-\sqrt{1-\frac{1}{\mathrm{x}}})=$$$$\mathrm{x}(\sqrt{\mathrm{x}-\frac{1}{\mathrm{x}}}-\sqrt{1-\frac{1}{\mathrm{x}}})$$$$\Rightarrow \mathrm{x}-\frac{1}{\mathrm{x}}-1+\frac{1}{\mathrm{x}}=$$$$\mathrm{x}(\sqrt{\mathrm{x}-\frac{1}{\mathrm{x}}}-\sqrt{1-\frac{1}{\mathrm{x}}})$$$$\Rightarrow \sqrt{\mathrm{x}-\frac{1}{\mathrm{x}}}-\sqrt{1-\frac{1}{\mathrm{x}}}=1-\frac{1}{\mathrm{x}}...\left(\mathrm{ii}\right)$$$$\left(\mathrm{i}\right)+\left(\mathrm{ii}\right)$$$$2\sqrt{\mathrm{x}-\frac{1}{\mathrm{x}}}=\mathrm{x}-\frac{1}{\mathrm{x}}+1$$$$\mathrm{Put}\sqrt{\mathrm{x}-\frac{1}{\mathrm{x}}}=\mathrm{t}$$$$\Rightarrow 2\mathrm{t}={\mathrm{t}}^2+1$$$$\Rightarrow {(\mathrm{t}-1)}^2=0$$$$\Rightarrow \mathrm{t}=1$$$$\Rightarrow \sqrt{\mathrm{x}-\frac{1}{\mathrm{x}}}=1$$$$\Rightarrow \mathrm{x}-\frac{1}{\mathrm{x}}=1$$$$\Rightarrow {\mathrm{x}}^2-\mathrm{x}-1=0$$$$\mathrm{x}=\frac{1\pm \sqrt{5}}{2}$$$$\mathrm{but}\mathrm{x}{\mathrm{can}}^{\prime }\mathrm{t}\mathrm{be}\mathrm{negative}\mathrm{for}\mathrm{real}\mathrm{numbers}$$$$\mathrm{of}\mathrm{the}\mathrm{equation}$$$$\mathrm{so}\mathrm{x}=\frac{1+\sqrt{5}}{2}$$

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