Solve-for-real-numbers-1-x-2x-2-1-2x-1-x-2-

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

Solve for real numbers :

\sqrt{1 - x} = {2 x}^{2} - 1 - 2 x \sqrt{1 - x^{2}}

Answered by MJS_new last updated on 11/Dec/21

squaring & transforming

x (1 - 4 (2 x^{2} - 1) \sqrt{1 - x^{2}}) = 0

but x \neq 0

4 (2 x^{2} - 1) \sqrt{1 - x^{2}} = 1

squaring & transforming

x^{6} - 2 x^{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

now {it}^{'} s easy

testing all solution we 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

thank you so much my dear Sir very nice