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Question Number 98761 by bemath last updated on 16/Jun/20

$$\sqrt{\mathrm{x}+\sqrt{\mathrm{x}}}-\sqrt{\mathrm{x}-\sqrt{\mathrm{x}}}=\mathrm{m}\sqrt{\frac{\mathrm{x}}{\mathrm{x}+\sqrt{\mathrm{x}}}}$$$$\mathrm{m}\mathrm{is}\mathrm{a}\mathrm{real}\mathrm{parameter}$$

Commented by MJS last updated on 16/Jun/20

$$\mathrm{if}x\in \mathbb{R}$$$$\Rightarrow x>0\wedge x-\sqrt{x}⩾0\Rightarrow x⩾1$$$$\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}=m\sqrt{\frac{x}{x+\sqrt{x}}}$$$$\sqrt{(1+\sqrt{x})\sqrt{x}}-\sqrt{(-1+\sqrt{x})\sqrt{x}}=m\sqrt{\frac{\sqrt{x}}{1+\sqrt{x}}}$$$$x⩾1\Rightarrow \mathrm{we}\mathrm{are}\mathrm{allowed}\mathrm{to}\mathrm{divide}\mathrm{by}\sqrt{x}$$$$\sqrt{1+\sqrt{x}}-\sqrt{-1+\sqrt{x}}=\frac{m}{\sqrt{1+\sqrt{x}}}$$$$1+\sqrt{x}-\sqrt{-1+\sqrt{x}}\times \sqrt{1+\sqrt{x}}=m$$$$-1+\sqrt{x}⩾0\wedge 1+\sqrt{x}⩾0$$$$1+\sqrt{x}-\sqrt{x-1}=m$$$$\Rightarrow 1<m⩽2$$$$\sqrt{x}-\sqrt{x-1}=m-1$$$$\mathrm{squaring}(\mathrm{both}\mathrm{sides}>0)$$$$2x-1-2\sqrt{x(x-1)}={(m-1)}^2$$$$2\sqrt{x(x-1)}=2x-1-{(m-1)}^2$$$$\mathrm{to}\mathrm{prove}:2x-1-{(m-1)}^2⩾0$$$$2x-1⩾{(m-1)}^2$$$$x⩾1\Rightarrow 2x-1⩾1$$$$\sqrt{2x-1}⩾m-1$$$$\sqrt{2x-1}+1⩾m$$$$\sqrt{2x-1}+1⩾2\wedge m⩽2\Rightarrow \mathrm{true}$$$$\mathrm{squaring}(\mathrm{both}\mathrm{sides}>0)$$$$4x(x-1)={(2x-1-{(m-1)}^2)}^2$$$$4{(m-1)}^{2}x={(m^2-2m+2)}^2$$$$x=\frac{{(m^2-2m+2)}^2}{4{(m-1)}^2}\wedge 1<m⩽2$$

Commented by john santu last updated on 16/Jun/20

$$\sqrt{\mathrm{x}+\sqrt{\mathrm{x}}}\{\sqrt{\mathrm{x}+\sqrt{\mathrm{x}}}-\sqrt{\mathrm{x}-\sqrt{\mathrm{x}}}\}=\mathrm{m}\sqrt{\mathrm{x}}$$$$\mathrm{x}+\sqrt{\mathrm{x}}-\sqrt{{\mathrm{x}}^2-\mathrm{x}}=\mathrm{m}\sqrt{\mathrm{x}}$$$$\mathrm{x}+(1-\mathrm{m})\sqrt{\mathrm{x}}=\sqrt{{\mathrm{x}}^2-\mathrm{x}}$$$$\mathrm{squaring}$$$${\mathrm{x}}^2+2\mathrm{x}\sqrt{\mathrm{x}}(1-\mathrm{m})+{(1-\mathrm{m})}^2\mathrm{x}={\mathrm{x}}^2-\mathrm{x}$$$$2\mathrm{x}\sqrt{\mathrm{x}}(1-\mathrm{m})+{(1-\mathrm{m})}^2\mathrm{x}+\mathrm{x}=0$$$$\mathrm{x}\{2\sqrt{\mathrm{x}}(1-\mathrm{m})+2-2\mathrm{m}+{\mathrm{m}}^2\}=0$$$$2\sqrt{\mathrm{x}}=\frac{{\mathrm{m}}^2-2\mathrm{m}+2}{\mathrm{m}-1}\Rightarrow \sqrt{\mathrm{x}}=\frac{{\mathrm{m}}^2-2\mathrm{m}+2}{2\mathrm{m}-2}$$$$\mathrm{x}={\left(\frac{{\mathrm{m}}^2-2\mathrm{m}+2}{2\mathrm{m}-2}\right)}^2$$

Commented by MJS last updated on 16/Jun/20

$$\mathrm{your}\mathrm{answer}\mathrm{is}\mathrm{only}\mathrm{true}\mathrm{for}m=2$$

Commented by john santu last updated on 16/Jun/20

$$\mathrm{typo}\mathrm{sir}.\mathrm{your}\mathrm{answer}\mathrm{and}\mathrm{my}\mathrm{answer}$$$$\mathrm{is}\mathrm{same}$$

Commented by MJS last updated on 16/Jun/20

$$\mathrm{I}\mathrm{see}.\mathrm{but}\mathrm{we}\mathrm{need}\mathrm{to}\mathrm{restrict}m\mathrm{like}\mathrm{I}\mathrm{showed}$$

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