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Question Number 163451 by HongKing last updated on 07/Jan/22

$$\mathrm{Solve}\mathrm{for}\mathrm{real}\mathrm{numbers}:$$$$\sqrt{1-\mathrm{x}}=1-{2\mathrm{x}}^2+2\mathrm{x}\sqrt{1-{\mathrm{x}}^2}$$

Commented by mr W last updated on 07/Jan/22

$$\sqrt{1-x}=(1-x^2)+2x\sqrt{1-x^2}+x^2$$$$\sqrt{1-x}={\left(\sqrt{1-x^2}\right)}^2+2x\sqrt{1-x^2}+x^2$$$$\sqrt{1-x}={(x+\sqrt{1-x^2})}^2$$$$\Rightarrow weseex=0$$

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