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Question Number 163451 by HongKing last updated on 07/Jan/22
Solve for real numbers:  (√(1 - x)) = 1 - 2x^2  + 2x (√(1 - x^2 ))
$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\sqrt{\mathrm{1}\:-\:\mathrm{x}}\:=\:\mathrm{1}\:-\:\mathrm{2x}^{\mathrm{2}} \:+\:\mathrm{2x}\:\sqrt{\mathrm{1}\:-\:\mathrm{x}^{\mathrm{2}} } \\ $$
Commented by mr W last updated on 07/Jan/22
(√(1−x))=(1−x^2 )+2x(√(1−x^2 ))+x^2   (√(1−x))=((√(1−x^2 )))^2 +2x(√(1−x^2 ))+x^2   (√(1−x))=(x+(√(1−x^2 )))^2   ⇒we see x=0
$$\sqrt{\mathrm{1}−{x}}=\left(\mathrm{1}−{x}^{\mathrm{2}} \right)+\mathrm{2}{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }+{x}^{\mathrm{2}} \\ $$$$\sqrt{\mathrm{1}−{x}}=\left(\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)^{\mathrm{2}} +\mathrm{2}{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }+{x}^{\mathrm{2}} \\ $$$$\sqrt{\mathrm{1}−{x}}=\left({x}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)^{\mathrm{2}} \\ $$$$\Rightarrow{we}\:{see}\:{x}=\mathrm{0} \\ $$

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