Solve 2xsqrt{1-x^2}=sqrt{1-(1-x)^2}

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

Solve for x

$$x=\frac{\sqrt{17}-1}{4}\approx 0.780776406$$

$$x=0$$

$$x=\frac{1}{2}=0.5$$

Solution Steps

Square both sides of the equation.

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

Expand \left(2x\sqrt{1-x^{2}}\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

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

Calculate \sqrt{1-x^{2}} to the power of 2 and get 1-x^{2}.

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

Use the distributive property to multiply 4x^{2} by 1-x^{2}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-x\right)^{2}.

$$4x^{2}-4x^{4}=\left(\sqrt{1-\left(1-2x+x^{2}\right)}\right)^{2}$$

To find the opposite of 1-2x+x^{2}, find the opposite of each term.

$$4x^{2}-4x^{4}=\left(\sqrt{1-1+2x-x^{2}}\right)^{2}$$

Subtract 1 from 1 to get 0.

$$4x^{2}-4x^{4}=\left(\sqrt{2x-x^{2}}\right)^{2}$$

Calculate \sqrt{2x-x^{2}} to the power of 2 and get 2x-x^{2}.

$$4x^{2}-4x^{4}=2x-x^{2}$$

Subtract 2x from both sides.

$$4x^{2}-4x^{4}-2x=-x^{2}$$

Add x^{2} to both sides.

$$4x^{2}-4x^{4}-2x+x^{2}=0$$

Combine 4x^{2} and x^{2} to get 5x^{2}.

$$5x^{2}-4x^{4}-2x=0$$

Substitute t for x^{2}.

$$-4t^{2}+5t-2=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute -4 for a, 5 for b, and -2 for c in the quadratic formula.

$$t=\frac{-5±\sqrt{5^{2}-4\left(-4\right)\left(-2\right)}}{-4\times 2}$$

Do the calculations.

$$t=\frac{-5±\sqrt{-7}}{-8}$$

Since the square root of a negative number is not defined in the real field, there are no solutions.

$$t\in \emptyset$$

Since t=x^{2}, the original equation does not have any solutions.

$$x\in \emptyset$$