Answer

Given

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

Rearrange the term

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

Apply perfect square formula

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

Let

t = 1-x
(1)

Then

t^2 = \sqrt{t}

Square both sides

t^4=t

Then

t(t^3-1) = 0

Then

t = 0

or

t^3-1 = 0

Apply difference of cubes formula

(t-1)(t^2+t+1) = 0

Then

t = 1

or

t^2+t+1 = 0

Solve the quadratic equation give two complex roots

t = -\dfrac{1}{2} \pm\dfrac{\sqrt{3} }{2}i

Noe we have four roots for t

\begin{cases} t_1 &=0 \\ t_2 &=1 \\ t_{3,4} &= -\dfrac{1}{2} \pm\dfrac{\sqrt{3} }{2}i \end{cases}

Substituting to (1) gives 4 solutions for x

\begin{cases} x_1 &=0 \\ x_2 &=1 \\ x_{3,4} &= \dfrac{3}{2} \pm\dfrac{\sqrt{3} }{2}i \end{cases}

Steven Zheng posted 5 hours ago

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