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Question Number 161091 by HongKing last updated on 11/Dec/21

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

Answered by MJS_new last updated on 11/Dec/21

$$\mathrm{squaring}\mathrm{\&}\mathrm{transforming}$$$$x(1-4(2x^2-1)\sqrt{1-x^2})=0$$$$\mathrm{but}x\ne 0$$$$4(2x^2-1)\sqrt{1-x^2}=1$$$$\mathrm{squaring}\mathrm{\&}\mathrm{transforming}$$$$x^6-2x^4+\frac{5}{2}{x}^2-\frac{15}{64}=0$$$$(x^2-\frac{3}{4})(x^2-\frac{5}{8}-\frac{\sqrt{5}}{8})(x^2-\frac{5}{8}+\frac{\sqrt{5}}{8})=0$$$$\mathrm{now}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{easy}$$$$\mathrm{testing}\mathrm{all}\mathrm{solution}\mathrm{we}\mathrm{get}$$$$x_1=-\frac{\sqrt{3}}{2}$$$$x_{2,3}=\pm \frac{\sqrt{10+2\sqrt{5}}}{4}$$

Commented by HongKing last updated on 11/Dec/21

$$\mathrm{thank}\mathrm{you}\mathrm{so}\mathrm{much}\mathrm{my}\mathrm{dear}\mathrm{Sir}\mathrm{very}\mathrm{nice}$$

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