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

Solve for real numbers:  (√(1 - x)) = 2x^2  - 1 - 2x (√(1 - x^2 ))

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

Answered by MJS_new last updated on 11/Dec/21

squaring & transforming  x(1−4(2x^2 −1)(√(1−x^2 )))=0  but x≠0  4(2x^2 −1)(√(1−x^2 ))=1  squaring & transforming  x^6 −2x^4 +(5/2)x^2 −((15)/(64))=0  (x^2 −(3/4))(x^2 −(5/8)−((√5)/8))(x^2 −(5/8)+((√5)/8))=0  now it′s easy  testing all solution we get  x_1 =−((√3)/2)  x_(2, 3) =±((√(10+2(√5)))/4)

$$\mathrm{squaring}\:\&\:\mathrm{transforming} \\ $$$${x}\left(\mathrm{1}−\mathrm{4}\left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}\right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)=\mathrm{0} \\ $$$$\mathrm{but}\:{x}\neq\mathrm{0} \\ $$$$\mathrm{4}\left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}\right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\mathrm{squaring}\:\&\:\mathrm{transforming} \\ $$$${x}^{\mathrm{6}} −\mathrm{2}{x}^{\mathrm{4}} +\frac{\mathrm{5}}{\mathrm{2}}{x}^{\mathrm{2}} −\frac{\mathrm{15}}{\mathrm{64}}=\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} −\frac{\mathrm{3}}{\mathrm{4}}\right)\left({x}^{\mathrm{2}} −\frac{\mathrm{5}}{\mathrm{8}}−\frac{\sqrt{\mathrm{5}}}{\mathrm{8}}\right)\left({x}^{\mathrm{2}} −\frac{\mathrm{5}}{\mathrm{8}}+\frac{\sqrt{\mathrm{5}}}{\mathrm{8}}\right)=\mathrm{0} \\ $$$$\mathrm{now}\:\mathrm{it}'\mathrm{s}\:\mathrm{easy} \\ $$$$\mathrm{testing}\:\mathrm{all}\:\mathrm{solution}\:\mathrm{we}\:\mathrm{get} \\ $$$${x}_{\mathrm{1}} =−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${x}_{\mathrm{2},\:\mathrm{3}} =\pm\frac{\sqrt{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{4}} \\ $$

Commented by HongKing last updated on 11/Dec/21

thank you so much my dear Sir very nice

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

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