Solve x=sqrt{4-2x}+2

$$x = \sqrt{ 4-2x } +2$$

Solve for x

$$x=2$$

Solution Steps

Subtract 2 from both sides of the equation.

$$x-2=\sqrt{4-2x}$$

Square both sides of the equation.

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

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

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

Calculate \sqrt{4-2x} to the power of 2 and get 4-2x.

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

Subtract 4 from both sides.

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

Subtract 4 from 4 to get 0.

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

Add 2x to both sides.

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

Combine -4x and 2x to get -2x.

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

Factor out x.

$$x\left(x-2\right)=0$$

To find equation solutions, solve x=0 and x-2=0.

$$x=0 x=2$$

Substitute 0 for x in the equation x=\sqrt{4-2x}+2.

$$0=\sqrt{4-2\times 0}+2$$

Simplify. The value x=0 does not satisfy the equation.

$$0=4$$

Substitute 2 for x in the equation x=\sqrt{4-2x}+2.

$$2=\sqrt{4-2\times 2}+2$$

Simplify. The value x=2 satisfies the equation.

$$2=2$$

Equation x-2=\sqrt{4-2x} has a unique solution.

$$x=2$$